Find the equation of a circle with Centre (a cos ∝, a sin ∝) and r

1.	Find the equation of a circle with Centre (a cos ∝, a sin ∝) and radius a
Answer» The general form of the equation of a circle is: (x - h)² + (y - k)² = r² Where, (h, k) is the centre of the circle. r is the radius of the circle. Substituting the centre and radius of the circle in he general form: (x - (a cos ∝))² + (y - (a sin ∝))² = a² ⇒ (x - a cos ∝)² + (y - a sin ∝)²= a² ⇒ x² - 2xacos α + a² cos² α + y²- 2yasin α + a² sin² α = a² ⇒ x² + y² + a² (cos² α + sin² α) - 2a(xcos α + ysin α) = a² ⇒ x² + y² + a² - 2a(xcos α + ysin α) = a² …((cos2α + sin²α) = 1) ⇒ x² + y² - 2a(xcos α + ysin α) = 0 equation of a circle with Centre (a cos ∝, a sin ∝) and radius a is: x² + y² - 2a(xcos α + ysin α) = 0

Find the equation of a circle with Centre (a cos ∝, a sin ∝) and radius a

Answer»

The general form of the equation of a circle is:

(x - h)² + (y - k)² = r²

Where, (h, k) is the centre of the circle.

r is the radius of the circle.

Substituting the centre and radius of the circle in he general form:

(x - (a cos ∝))² + (y - (a sin ∝))² = a²

⇒ (x - a cos ∝)² + (y - a sin ∝)²= a²

⇒ x² - 2xacos α + a² cos² α + y²- 2yasin α + a² sin² α = a²

⇒ x² + y² + a² (cos² α + sin² α) - 2a(xcos α + ysin α) = a²

⇒ x² + y² + a² - 2a(xcos α + ysin α) = a² …((cos2α + sin²α) = 1)

⇒ x² + y² - 2a(xcos α + ysin α) = 0

equation of a circle with Centre (a cos ∝, a sin ∝) and radius a is:

x² + y² - 2a(xcos α + ysin α) = 0

Discussion

No Comment Found

Related InterviewSolutions

The loucs of the centre of the circle which cuts orthogonally the circle `x^(2)+y^(2)-20x+4=0` and which touches x=2 isA. `x^(2)=16y`B. `x^(2)=16y+4`C. `y^(2)=16x`D. `y^(2)=16x+4`
If a variable line, `3x + 4y -lambda = 0` is such that the two circles `x^(2)+y^(2)-2x-2y+1=0` and `x^(2)+y^(2)-18x-2y+78=0` are not its opposite sides, then the set of all values of `lambda` is the intervalA. [13, 23]B. [2,17]C. [12,21]D. [23,31]
The equation of a circle which cuts the three circles `x^(2)+y^(2)+2x+4y+1=0,x^(2)+y^(2)-x-4y+8=0` and `x^(2)+y^(2)+2x-6y+9=0` orthogonlly isA. `x^(2)+y^(2)-2x-4y+1=0`B. `x^(2)+y^(2)+2x+4y+1=0`C. `x^(2)+y^(2)-2x+4y+1=0`D. `x^(2)+y^(2)-2x-4y-1=0`
Two variable chords AB and BC of a circle `x^(2)+y^(2)=a^(2)` are such that `AB=BC=a`. M and N are the midpoints of AB and BC, respectively, such that the line joining MN intersects the circles at P and Q, where P is closer to AB and O is the center of the circle. The locus of the points of intersection of tangents at A and C isA. `x^(2)+y^(2)=a^(2)`B. `x^(2)+y^(2)=2a^(2)`C. `x^(2)+y^(2)=4a^(2)`D. `x^(2)+y^(2)=8a^(2)`
One of the limiting points of the co-axial system of circles containing the circles `x^(2)+y^(2)-4=0andx^(2)+y^(2)-x-y=0` isA. `(sqrt2,sqrt2)`B. `(-sqrt2,sqrt2)`C. `(-sqrt2-sqrt2)`D. None of these
The limiting points of the system of circles represented by the equation `2(x^(2)+y^(2))+lambda x+(9)/(2)=0`, areA. `(pm(3)/(2),0)`B. `(0,0)and((9)/(2),0)`C. `(pm(9)/(2),0)`D. `(pm3,0)`
`t_(1),t_(2),t_(3)` are lengths of tangents drawn from a point (h,k) to the circles `x^(2)+y^(2)=4,x^(2)+y^(2)-4=0andx^(2)+y^(2)-4y=0` respectively further, `t_(1)^(4)=t_(2)^(2)" "t_(3)^(2)+16`. Locus of the point (h,k) consist of a straight line `L_(1)` and a circle `C_(1)` passing through origin. A circle `C_(2)` , which is equal to circle `C_(1)` is drawn touching the line `L_(1)` and the circle `C_(1)` externally. Equation of `L_(1)` isA. x+y=0B. x-y=0C. 2x+y=0D. x+2y=0
Find the equation of the radical axis of circles `x^2+y^2+x-y+2=0` and `3x^2+3y^2-4x-12=0`A. `2x^(2)+2y^(2)-5x+y-14=0`B. `7x-3y+18=0`C. `5x-y+14=0`D. None of these
Find the equation of radical axis of the circles `x^(2)+y^(2)-3x+5y-7=0` and `2x^(2)+2y^(2)-4x+8y-13=0`.
The radius centre of the circles `x^(2)+y^(2)=1,x^(2)+y^(2)+10y+24=0andx^(2)+y^(2)-8x+15=0` isA. (2,5/2)B. (-2,5/2)C. (-2,-5/2)D. (2,-5/2)