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Find the coordinates of the point at which the circles `x^2-y^2-4x-2y+4=0` and `x^2+y^2-12 x-8y+36=0` touch each other. Also, find equations of common tangents touching the circles the distinct points. |
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Answer» Correct Answer - `y = 0 and 7y -24x + 16=0` two circles touch each other externally, if `C_(1)C_(2) = r_(1)+r_(2)` and internally if `C_(1)C_(2)= r_(1)~r_(2)` Given circles are `x^(2)+y^(2) -2y+4=0`, whose centre `C_(1)(2, 1)` and radius `r_(1)=1` and `x^(2) +y^(2)- 12x -8y +36=0` whose centre `C_(2)(6,4)` and radius `r_(2)=4` The distance between the centres is `sqrt((6-2)^(2)+(4-1)^(2))=sqrt(16+9)=5` `rArr C_(1) C_(2) =r_(1)+r_(2)` Therefore, the circles touch each other externally and at the point of touching the point divides the line joining the two centres internally in the ratio of their radii, `1 : 4` Therefore, `x_(1) = (1xx6+4xx2)/(1+4)=(14)/(5)` `y_(1)= (1xx4+4xx1)/(1+4)=(8)/(5)` Again, to determine the equation of common tangents touching the circles in distinct points, we know that, the tangents pass through a point which divides the line joining the two centres externally in the ratio of their radi, i.e. `1:4` Therefore, `x_(2)=(1xx6-4xx2)/(1-4)=(-2)/(-3)=(2)/(3)` `and y_(2)=(1xx4-4xx1)/(1-4)=0` Now, let m be the slope of the tangent and this line passing through (2/3, 0 ) is `y-0=m(x-2//3)` `y-mx+(2)/(3)m=0` This is tangent to the Ist circle, if perpendicular distance from centre = radius. `therefore(1-2m+(2//3)m)/(sqrt(1+m^(2)))=1 [therefore C_(1)=(2,1) and r_(1)=1]` `rArr 1-(4)/(3)m=sqrt(1+m^(2))` `rArr1+(16)/(9)m^(2)-(8)/(3)m=1+m^(2)` `rArr (7)/(3)m^(2)-(8)/(3)m=0` `rArr m((7)/(9)m-(8)/3)=0` `rArr m = 0 ,m=(27)/(7)` Hence, the equation of two tangents are `y=0and y=(27)/(7)(x-(2)/(3))` `rArr y = 0 and 7y -24x + 16 = 0` |
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