Question

Equation of the circle which passes through the point $(-1,2)$ and touches the circle $x^2+y^2-8x+6y=0$ at origin is

• A. $x^2+y^2-2x-\frac{3}{2}y=0$
• B. $x^2+y^2-x-2y=0$
• C. $x^2+y^2+2x+\frac{3}{2}y=0$
• D. $x^2+y^2+2x-\frac{3}{2}y=0$

Answer 1

The correct option is D $x^2+y^2+2x-\frac{3}{2}y=0$

Since, circles are tangents at point $(0,0)$, then line that passes through centres of both circles must go through origin.

Line through $(0,0)$ and $(4,-3)$ has equation

$y=\frac{-3}{4}x4y+3k=0$

Centre $(h,k)$ must lies on this line

$4k+3h=0\left(1\right)$

Distance of $(h,k)$ from $(-1,2)$ and $(0,0)$ must be equal to radius

$(h+1{)}^2+(y-2{)}^2=(h-0{)}^2+(y-0{)}^2{h}^2+k^2+2h-4k+5=h^2+k^{2}2h-4k=-5\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$h=-1,k=\frac{3}{4}r=h^2+k^2=\frac{5}{4}(x-1{)}^2+{(y+\frac{3}{4})}^2=\frac{5}{4}{x}^2-2x+y^2+\frac{3}{2}y=0$