435 + Interview Questions in CIRCLE IN GENERAL AWARENESS Page 5 InterviewSolution

201.	In the adjoining figure, `O` is the centre of the circle. If the chord AB is equal to the radius of the circle, then find the value of `angleADB`.
Answer» Correct Answer - `30^@`

Discussion

202.	A rectangle in inscribed in a circle of radius 5 cm. if the breadth of the rectangle is 6 cm. then find the length of the rectangle.
Answer» Correct Answer - 8 cm

Discussion

203.	In the adjoining figure, ABCD is a cyclic quadrilateral . If `angleCBX=104^(@)` and `angleCAB=56^(@)`, then find the value of `angleADB`.
Answer» square ` ABCD` is a cyclic quadrilateral. `therefore angle ADC=angle CBX=104^@` and `angle BDC=angle CAB` (angles of same segment) `rArrangleBDC=56^@` `therefore angle ADB=angle ADC-angle BDC` `=104^@-56^@` `48^@`.

Discussion

204.	In the adjoining figure, `O` is the centre of the circle and AB is its diameter. If AC=8 cm and BC= 6cm, then find the radius of the circle.
Answer» Correct Answer - 5 cm

Discussion

205.	In the adjoining figure, AB is a chord of the circle. If `angleAEB=110^@` and `angleEBC=25^@`,then find the value of `angleADB`.
Answer» Correct Answer - `85^@`

Discussion

206.	In the adjoining figure, `O` is the centre of the circle. Find the value of `angle BEC`.
Answer» Correct Answer - `105^@`

Discussion

207.	In the adjoining figure,AB is the diameter of the circle of centre `O` & the chord CD is equal to radius. If P is an external point, then find the value of `angleAPB`.
Answer» Correct Answer - `60^@`

Discussion

208.	In the adjoining figure,`O` is the centre of the circle. If `angleABC=110^@` , then find the value of `angleAOC`.
Answer» Correct Answer - `140^@`

Discussion

209.	In the adjoining figure, AB is the diameter of the circle and two points C and D are on the circle. If `angle CAD=45^@` and `angleABC=65^@`, then find the value of `angle DCA`.
Answer» Correct Answer - `20^@`

Discussion

210.	`O` is the centre of a circle and an equilateral `DeltaABC` is inscribed in it. Find the value of `angleBOC`.
Answer» Correct Answer - `120^@`

Discussion

211.	What happen to area of circle, if its radius is doubled?
Answer» Correct Answer - It becomes 4 times. Values : If we increase our moral values we shall get more respect in our society.

Discussion

212.	Observe the following figure and complete the following activity `:` `/_ ABC " is " 90^@`
Answer» Seg AC is the diameter `:. M = (arc AMC ) = 180^(@)` `/_ ABC = (1)/(2) m ( arc square ) ` …..(Inscribed angle theorem ) `:. /_ABC = (1)/(2) xx square ` `:. /_ABC = square ` `:. ` Angle inscribed in a semicircle is a `square `. Activity `:` Seg AC is the diameter `:. `m ( arc AMC ) `= 180^(@)` `/_ ABC = (1)/(2) m (arc AMC ) ` ...(Inscribed angle theorem ) `:. /_ABC = (1)/(2) xx 180^(@)` `:. /_ABC = 90^(@)` ` :.` Angle inscribed in a semicircle is a right angle.

Discussion

213.	In the figure, chord AB `~= ` chord CD Prove that , ar AC `~=` arc BD
Answer» Chord `AB ~=` chord CD ….(Given ) `:.` Arc ACB `~=` arc CBD …(Arcs corresponding to congruent chords ) `:.` m (arc ACB ) = m (arc CBD ) ….(1) But m ( arc ACB ) = m ( arc AC ) + m ( arc CB ) …..(2) and m (arc CBD ) = m ( arc CB ) + m (arc BD ) ....(Arc addition postulate ) ....(3) From (1 , (2) and (3) , we get m (arc AC + m (arc CB ) = m (arc CB ) + m (arc BD ) `:. ` m (arc AC ) = m ( arc BD ) `:. ` arc `~=`` arc BD.

Discussion

214.	Points A, B, C are on a circle, such that m(arc AB) = m (arc BC) `=120^(@)` . No point, except point B, is common to arcs. Which is the type of `DeltaABC` ?A. Equilateral triangleB. Scalene triangleC. Right angled triangleD. Isosceles triangle
Answer» Correct Answer - A

Discussion

215.	The length of the longest chord of the circle with radius 2.9 cm is ____. (A) 3.5 cm (B) 7 cm (C) 10 cm (D) 5.8 cm
Answer» (D) 5.8 cm Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm

215.

The length of the longest chord of the circle with radius 2.9 cm is ____. (A) 3.5 cm (B) 7 cm (C) 10 cm (D) 5.8 cm

Answer»

(D) 5.8 cm

Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm

Discussion

216.	Find the equation of a circle passing through the points `(5, 7), (6,6) and (2, -2)`. Find its centre and radius.
Answer» Let the required equation of the circle be ` x ^(2) + y ^(2) + 2gx + 2 fy + c = 0 " "`… (i) Since it passes through each of the points ` (5, 7 ), (6, 6 ) and (2, - 2 )`, each one of these points must satisfy (i). ` therefore " " 25 + 49 + 10 g + 14 f +c= 0 rArr 10 g + 14 f + c + 74 = 0" "` ... (ii) ` " " 36 + 36 + 12 g + 12 f + c = 0 rArr 12 g + 12 f + c + 72 = 0 " " `... (iii) ` " " 4 + 4 + 4 g - 4 f + c = 0 " " rArr 4 g - 4 f + c + 8 = 0 " " ` ... (iv) Subtracting (ii) from (iii), we get ` 2g - 2f - 2 = 0 rArr g - f = 1 " "` ... (v) Subtracting (iv) from (iii), we get ` 8g + 16 f + 64 = 0 " " rArr g + 2f = - 8" " `... (vi) Solving (v) and (vi), we get ` g = - 2 and f = - 3` Putting ` g = - 2 and f = - 3 ` in (ii), we get ` c = -12 ` Putting ` g = - 2 , f = - 3 and c = - 12 ` in (i) , we get ` x ^(2)+ y ^(2) - 4x - 6y - 12=0`, which is the required equation of the circle. Centre of this circle ` = (-g, - f ) = (2, 3 )` And, its radius =` sqrt (g ^(2) + f ^(2) - c ) = sqrt ( 4 + 9 + 12 ) = sqrt ( 25) = 5` units

Discussion

217.	Find the equation of the circle passing through the centre of the circle `x^2 + y^2 + 8x + 10 y - 7 =0` and concentric with the circle `x^2 + y^2 - 4x - 6y = 0`.
Answer» We have ` 2x ^(2) + 2y ^(2) - 8x - 12 y - 9 =0 ` `rArr x ^(2) + y ^(2) - 4x - 6y - (9)/(2) = 0 rArr x ^(2) + y ^(2) + 2gx + 2 f y + c = 0 ` ` " "` where ` g = - 2 , f =-3 and c = (-9)/(2)`. Centre of this circle ` = (-g, - f )= (2,3 )` ` therefore ` the centre of the required circle = `C(2, 3 )` Again , ` x ^(2) + y ^(2) + 2 gx + 2 fy + c = 0, ` where ` g = 4, f = 5 and c = - 7` Centre of this circle = `(-g, - f )= (- 4, - 5)`. So, the required circle passes through the point ` P(-4, - 5)`. Radius of the required circle ` = CP = sqrt (( 2+ 4 ) ^(2) + ( 3+ 5 ) ^(2)) ` `" "= sqrt ( 36 + 64 ) = sqrt (100) = 10 ` Hence, the required equation of the circle is ` (x - 2 ) ^(2) + (y - 3 ) ^(2) = (10)^(2) rArr x ^(2) + y ^(2) - 4x - 6y - 87 = 0 `

Discussion

218.	If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of `70^(@)`, then `/_POA` is equal toA. `70^(@)`B. `55^(@)`C. `100^(@)`D. `40^(@)`
Answer» Correct Answer - B

Discussion

219.	If TP and TQ are two tangents to a circle with centre O, so that `/_POQ = 120^(@)`, then `/_PTQ ` is equal toA. `120^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)`
Answer» Correct Answer - C

Discussion

220.	Find the number of point `(x ,y)`having integral coordinates satisfying the condition `x^2+y^2
Answer» Correct Answer - A

Discussion

221.	Two circle `x^2+y^2=6`and `x^2+y^2-6x+8=0`are given. Then the equation of the circle through their points ofintersection and the point (1, 1) is`x^2+y^2-6x+4=0``x^2+y^2-3x+1=0``x^2+y^2-4y+2=0`none of theseA. `x^(2)+y^(2)-6x+4=0`B. `x^(2)+y^(2)-3x+1=0`C. `x^(2)+y^(2)-4x+2=0`D. `x^(2)+y^(2)-2x+1=0`
Answer» Correct Answer - B

Discussion

222.	The two circles `x^2+ y^2=r^2` and `x^2+y^2-10x +16=0` intersect at two distinct points. ThenA. `rlt2`B. `rgt8`C. `2gtrlt8`D. `2lerle8`
Answer» Correct Answer - C

Discussion

223.	If the circle `x^2+y^2 +4x+22y + c = 0` bisects the circumference of the circle `x^2 + y^2-2x + 8y-d = 0`,then `(c+d)`is equal toA. 40B. 50C. 60D. 70
Answer» Correct Answer - B

Discussion

224.	Find the values of `k`for which the points `(2k ,3k),(1,0),(0,1),a n d(0,0)`lie on a circle.
Answer» Correct Answer - `5//13` The circle passing through (1,0),(0,1) and (0,0) is `x(x-1)+y(y-1)=0` It also passes through the point (2k,3k). Therefore, `2k(2k-1)+3k(3k-1)=0` i.e., `k=0` or `k=(5)/(13)`

Discussion

225.	Find the equation of the circle passing through the origin and cuttingintercepts of lengths 3 units and 4 unitss from the positive exes.
Answer» Correct Answer - `x^(2)+y^(2)-3x-4y=0` Obviously, (3,0) and (0,4) are the endpoints of diameter. Then , the equation is `(x-3)(x-0)+(y-0)(y-4)=0 ` i.e., `x^(2)+y^(2)-3x-4y=0`

Discussion

226.	Prove that the circle drawn with any side of a rhombus as a diameter,posses through the point of intersection of its diagonals.
Answer» According to diagram `/_AEB=90^@` Circle with AB as diameter passes through E.

Discussion

227.	The angle of which the circle `(x-1)^2+y^2=10 and x^2+(y-2)^2=5` intersect is
Answer» Here, `C_(1)-=(1,0` and `r_(1)=sqrt(10)`. And `C_(2) -= (0,2)` and `r_(2)=sqrt(5)`. Also, `C_(1)C_(2)=sqrt(1^(2)+2^(2))=sqrt(5)` If circles intersect at an angle `theta`, then `cos theta =((C_(1)C_(2))^(2)-r_(1)^(2)r_(2)^(2))/(2r_(1).r_(2))=(5-10-5)/(2.sqrt(10).sqrt(5))=-(1)/(sqrt(2))` `:. theta =(3pi)/(4)`

Discussion

228.	If the circles `x^2+y^2+2x+2ky+6=0` and `x^2+y^2+2ky+k=0` intersect orthogonally then k equals (A) `2 or -3/2` (B) `-2 or -3/2` (C) `2 or 3/2` (D) `-2 or 3/2`A. 2 or -3/2B. `-2 or -3//2`C. 2 or 3/2D. `-2 or 3//2`
Answer» Correct Answer - A Since, the given circles intersect orthogonally. `therefore 2(1) (0)+2(k)(k)=6+k` `[therefore2g_(1)+2f_(1)f_(2)=c_(1)+c_(2)]` `rArr 2k^(2)-k-6=0rArrk=-(3)/(2),2`

Discussion

229.	If a point P is moving such that the length of tangents drawn from P to `x^(2) + y^(2) - 2x + 4 y - 20= 0" ___"(1)`. and `x^(2) + y^(2) - 2x - 8 y +1= 0" ___"(2)`. are in the ratio `2:1` Then show that the equation of the locus of P is `x^(2) + y^(2) - 2x - 12 y +8= 0`A.B.C.D.
Answer» Correct Answer - `x^(2) y^(2) - 2x -12y + 8 = 0`

Discussion

230.	Match the conics in List I with the statements`//` expressions in List II.
Answer» Correct Answer - `r rarr p` `(\|0+0-1\|)/(sqrt(h^(2)+k^(2)))=2` `implies 1 =4(k^(2)+h^(2))` `:. h^(2)+k^(2)=((1)/(2))^(2)` which is a circle. Note `:` Solutions of the remaining parts are given in their repective chapters.

Discussion

231.	Let `S`be the circle in the `x y`-plane defined by the equation `x^2+y^2=4.`(For Ques. No 15 and 16)Let `P`be a point on the circle `S`with both coordinates being positive. Let the tangentto `S`at `P`intersect the coordinate axes at the points `M`and `N`. Then, the mid-point of the line segment `M N`must lie on the curve`(x+y)^2=3x y`(b) `x^(2//3)+y^(2//3)=2^(4//3)`(c) `x^2+y^2=2x y`(d) `x^2+y^2=x^2y^2`A. `(x+y)^(2) = 3xy`B. `x^(2//3)+y^(2//3) =2^(4//3)`C. `x^(2)+y^(2)=2xy`D. `x^(2)+y^(2)=x^(2)y^(2)`
Answer» Correct Answer - 4 Let the coordinates of P be `(2 cos theta , 2 sin theta )` Equation of tangent to circle at P is `x cos x + y sin theta =2` `:. M((2)/(cos theta ,0)),N(0,(2)/(cos theta))` Let the mid -point of MN be `(x,y)` `:. x= (1)/(cos theta ) ` and `y = (1)/(sin theta)` Squaring and adding , we get `(1)/(x^(2))+(1)/(y^(2))=1 ` or `x^(2)+y^(2)=x^(2)y^(2)`, this is the required locus

Discussion

232.	A family of circles passing through the points `(3, 7) and (6, 5)` cut the circle `x^2 + y^2 - 4x-6y-3=0`. Show that the lines joining the intersection points pass through a fixed point and find the coordinates of the point.
Answer» Correct Answer - `x = 2 and y = 23//3` The equation of the circle on the line joining the points A (3, 7) and B (6, 5) as diameter is `(x-3)(x-6)+(y-7)(y-5)=0" "...(i)` and the equation of the line joining the point A (3, 7) and B (6, 5) is `y-7=(7-6)/(3-6)(x-3)` `rArr 2x + 3y - 27 = 0" "...(ii)` Now, the equation of family of circles passing through the point of intersection of Eqs. (i) and (ii) is `S+lambdaP=0` `rArr (x-3)( x-6)+(y-7)(y-5)+lambda(2x+3y-27)=0` `rArr x^(2)-6x-3x+18+y^(2)-5-7y+35` `+2lambdax+3lambday-27lambda=0` `rArr S_(1)-=x^(2)+y^(2)+x(2lambda-9)+y(3lambda-12)` `+(53-27lambda)=0" "...(iii)` Again, the circle,which cut the members of family of circles, is `S_(2)-=x^(2)+y^(2)-4x-6y-3=0" "...(iv)` and the equation of common chord to circles `S_(1) and S_(2)` is `S_(1)-S_(2)=0` `rArrx(2lambda-9+4)+y(3lambda-12+6)+(53-27lambda+3)=0` `rArr 2lambdax - 5x +3lambday-6y+56-27lambda=0` `rArr (-5x-6y+56)+lambda(2x+3y-27)=0` which represents equations of two straight lines passing through the fixed point whose coordinates are obtained by solving the two equations 5x + 6y - 56 = 0 and 2x + 3y - 27 = 0 we get x = 2 and y = 23/3

Discussion

233.	Consider the family ol circles `x^2+y^2=r^2, 2 < r < 5` . If in the first quadrant, the common tangnet to a circle of this family and the ellipse `4x^2 +25y^2=100` meets the co-ordinate axes at A and B, then find the equation of the locus of the mid-point of AB.
Answer» Correct Answer - `4x^(2)+25y^(2)=4x^(2)y^(2)` Equation of any tangent to circle `x^(2)+y^(2)=r^(2)` is `xcostheta + y sintheta = r" "...(i)` Suppose Eq. (i) is tangent to `4x^(2) + 25y^(2)=100` or when `(x^(2))/(25)+(y^(2))/(4)=1 " at"(x_(1),y_(2))` Then, Eq, (i) and `("xx"_(1))/(25)+(yy_(1))/(4)=1` are identical `therefore (x_(1)//25)/(costheta)=((y_(1))/(4))/(sintheta)=(1)/(r)` `rArr x_(1)(25costheta)/(r),y_(1)=(4sintheta)/(r)` The line (i) meet the coordinates axes in A `(rsectheta, 0) and beta(0,r cosec theta)`. LEt (h,k) be mid-point of AB. Then , " " `h=(rsectheta)/(2)and k=(rcosectheta)/(2)` Therefore, `2h=(r)/(costheta)and 2k=(r)/(sintheta)` `therefore x_(1)=(25)/(2h)and y_(1)=(4)/(2k)` As `(x_(1), y_(1)) " lies on the ellipse " (x^(2))/(25)+(y^(2))/(4)=1` we get `(1)/(25)((625)/(4h^(2)))+(1)/(4)((4)/(k^(2)))=1` `rArr (25)/(4h^(2))+(1)/(h^(2))=1` or `25k^(2)+4h^(2)=4h^(2)k^(2)` Therefore, required locus is `4x^(2)+25y^(2)=4x^(2)y^(2)`

Discussion

234.	The points of intersection of the line `4x-3y-10=0`and the circle `x^2+y^2-2x+4y-20=0`are ________ and ________
Answer» Correct Answer - `(-2, - 6) and (4, 2)` For point of intersection, we put `x=(3y+10)/(4)" in " x^(2)+y^(2)-2x+4y-20=0` `rArr ((3y+10)/(4))^(2)+y^(2)-2((3y+10)/(4))+4y-20=0` `rArr 25y^(2)+100y-300=0` `rArr y^(2)+4y-12=0` `rArr (y-2)(y+6)=0` `rArr y = -6 ,2` when y = -6 `rArr` x = -2 when y = 2 `rArr x = 4 `therefore` point intersection are (-2, -6) and (4, 2).

Discussion

235.	A circle is of radius 5 cm. Its longest chord will be :(A) 5 cm(B) 8 cm(C) 10 cm(D) 15 cm
Answer» Answer is (C) 10 cm Diameter = longest chord = 2 × radius = 2 × 5 = 10 cm

235.

A circle is of radius 5 cm. Its longest chord will be :(A) 5 cm(B) 8 cm(C) 10 cm(D) 15 cm

Answer»

Answer is (C) 10 cm

Diameter = longest chord

= 2 × radius = 2 × 5

= 10 cm

Discussion

236.	If in given fig. ΔABC is an equilateral triangle, then ∠BCD will be :(A) 140°(B) 120°(C) 180°(D) 275°
Answer» Answer is (B) 120° In ΔABC ∠A = 60° and in cyclic quadrilateral ABCD ∠A + ∠D = 180° ∠D = 180° – ∠A = 180° – 60° ∠BDC = 120°

236.

If in given fig. ΔABC is an equilateral triangle, then ∠BCD will be :(A) 140°(B) 120°(C) 180°(D) 275°

Answer»

Answer is (B) 120°

In ΔABC ∠A = 60° and in cyclic quadrilateral ABCD

∠A + ∠D = 180°

∠D = 180° – ∠A

= 180° – 60°

∠BDC = 120°

Discussion

237.	If a straight line through `C(sqrt(-8),sqrt(8))` make an angle `135^@` with the x-axis , cuts the circle `x=5costheta, y=5sintheta` in points A and B , find length of segment AB.A. 3B. 5C. 8D. 10
Answer» Correct Answer - D

Discussion

238.	In the figure, `/_PQR = 85^(@)` then find the measure of `/_ PTR ` . A. `75^(@)`B. `180^(@)`C. `90^(@)`D. `85^(@)`
Answer» Correct Answer - D

Discussion

239.	Find the equation of the smallest circle passing through theintersection of the line `x+y=1`and the circle `x^2+y^2=9`
Answer» Let the given circle and line intersect at points A and B. Equation of family of cirlces through A and B is `x^(2)+y^(2)-9+lambda(x+y-1)=0,lambda in R` Variable centre of the circle is `(-(lambda)/(2),-(lambda)/(2))`. The smallest circle of this family is that for which A and B are end points of diameter. So, centre `(-(lambda)/(2),-(lambda)/(2))`. lies ont he chord `x+y-1=0`. `implies -(lambda)/(2),-(lambda)/(2)-1=0` `implies lambda= -1` Using this value for `lambda`, the equation of the smallest circle is `x^(2)+y^(2)-9-(x+y-1)=0` or `x^(2)+y^(2)-x-y-8=0`

Discussion

240.	`angleXYZ=40^(@) ,angleAYZ=20^(@)` line Ay is tangent at point Y . `:.` m(arc XY) = ...... . A. `80^(@)`B. `40^(@)`C. `60^(@)`D. `120^(@)`
Answer» Correct Answer - C

Discussion

241.	AB is tangent at B. AB = 12, AP = 6 `:. PQ=` ....... A. 18B. 6C. 12D. 20
Answer» Correct Answer - A

Discussion

242.	If the circle `x^2+y^2=1`is completely contained in the circle `x^2+y^2+4x+3y+k=0`, then find the values of `kdot`
Answer» Given circles are `x^(2)+y^(2)=1` (1) and `x^(2)+y^(2)+4x+3y+k=0` (2) `C_(1)(0,0),r_(1)=1` `C_(2)(-2,-3//2),r_(2)=sqrt(4+(9)/(4)-k)=sqrt((25)/(4)-k)`. Circle (1) is completely contained by circle (2). `implies C_(1)C_(2) lt r_(2) -r_(1)` `implies sqrt(4+(9)/(4))ltsqrt((25)/(4)-k)` `implies (5)/(2)+1ltsqrt((25)/(4)-k)` `implies (25)/(4)-k gt(49)/(4)` `implies k lt - 6` Also , for these values of `k, (25)/(4)-k gt 0`.

Discussion

243.	The equation of the cirele which passes through the point (1, 1) and touches the circle `x^2+y^2+4x-6y-3=0` at the point (2, 3) on it is
Answer» Given circle is `x^(2)+y^(2)+4x-6y-3=0` (1) Required circle touches above circle at point P(2,3). So, it touches the tangent at point P to the given circle. Equation of tangent at point P(2,3) to the given circle is `2x+3y+2(x+2)-3(y+3)-3=0` or `x-2=0` Equatiion of family of circle touching line `x-2=0` at (2,3) is given by `[(x-2)^(2)+(y-3)^(2)]+lambda(x-2)=0,lambda in R` (2) It if passes through (1,1), then `1+4+lambda(-1)=0` or `lambda=5` Putting the value of `lambda` in equation (2), we get `(x-2)^(2)+(y-3)^(2)+5(x-3)=0` or `x^(2)+y^(2)+x-6y+3=0` This is the required circle.

Discussion

244.	`angleQPR=60^(@)` `:. angleAOB=` ........ A. `60^(@)`B. `90^(@)`C. `120^(@)`D. `240^(@)`
Answer» Correct Answer - C

Discussion

245.	Point P is on the circle. AB is diameter of the circle, `angleAPB` is .......A. a Reflex angleB. an acute angleC. a Right angleD. an obtuse angle
Answer» Correct Answer - A

Discussion

246.	Consider the circle `x^2+y^2-10x-6y+30=0`. Let O be the centre of the circle and tangent at A(7,3) and B(5, 1) meet at C. Let S=0 represents family of circles passing through A and B, thenA. the area of quadrilateral OACB is 4B. the radical axis for the famil of circles of `S=0` is `x+y=0`C. the smallest possible circle of the family `S=0` is `x+y-12x-4+38=0`D. the coordinates of point C are (7,1)
Answer» Correct Answer - 1,3,4 The coordinates of O are (5,3) and the radius is 2. The equation of tangent at A(7,3) is `7x+3y-5(x+7) -3(y+3) +30=0` `i.e., 2x-14=0` `i.e., x=7` The equation of tangent at `B (5,1)` is `5x+y-5(x+5)-3(y+1)+30=0` i.e., `-2y+2=0` i.e., `y=1` Therefore, the coordinates of C are (7,1) . So, area of OACB `=4` The equation of AB is `x-y=4` (radical axis). The equaiont of the smallest circle is `(x-7)(x-5)+(y-3)(y-1)=0` i.e., `x^(2)+y^(2)-12x-4y+38=0`

Discussion

247.	Tangent drawn from the point `(a ,3)`to the circle `2x^2+2y^2-25`will be perpendicular to each other if `alpha`equals5 (b)`-4`(c) 4(d) `-5`A. 5B. `-4`C. 4D. `-5`
Answer» Correct Answer - 2,3 The equation of the director circle is `x^(2)+y^(2)=25` . The point `(alpha,3)` must lie on this circle. So, `alpha^(2)+9=25` or `alpha^(2)= 16` or `alpha = +-4`

Discussion

248.	If `3x+y=0`is a tangent to a circle whose center is `(2,-1)`, then find the equation of the other tangent to the circle from theorigin.
Answer» Correct Answer - `x-3y=0` The angle between `3x+y=0` and the line joining (2,-1) to (0,0) is `theta =tan^(-1)\|(-3+(1)/(2))/(1+(-3)(-(1)/(2)))\|` `= tan^(-1)\|-1\|=(pi)/(4)` The other tangent is perpendicular to `3x+y=0` Therefore, the equation of the other tangent is `x-3y=0`.

Discussion

249.	An infinite number of tangents can be drawn from `(1,2)`to the circle `x^2+y^2-2x-4y+lambda=0`. Then find the value of `lambda`
Answer» Correct Answer - 5 Clearly, the point (1,2) is the center of the given circle and infinite tanegents can only be drawn on a point circle. Hence, the radius should be 0. Therefore, `sqrt(1^(2)+2^(2)-lambda)=0` or `lambda =5`

Discussion

250.	Find the area of the triangle formed by the tangents from the point (4,3) to the circle `x^2+y^2=9`and the line joining their points of contact.
Answer» Area of triangle formed by the tangents from (4, 3) to the circle `x^(2)+y^(2) = a^(2)` and their chord of contact `=a((h^(2)+k^(2)-a^(2))^(3//2))/(4^(2)+3^(2))=(3(16+9-9)^(3//2))/(25)` `=(3(64))/(25)=(192)/(25)`sq units.

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