Question

The tangent to the circle $x^2+y^2=5$ at the point $(1,-2)$ touches the circle $x^2+y^2-8x+6y+20=0$ at the point

• A. $(3,1)$
• B. $(-3,1)$
• C. $(3,-1)$
• D. $(-3,1-)$

Answer 1

The correct option is C $(3,-1)$

The tangents at $(1,-2)$ on the circle $x^2+y^2-5=0$ is $S_1=0,x-2y-5=0$
It touches the second circle at $(x_1,y_1)$
$\therefore x_1-2y_1-5=0$ ............$\left(1\right)$
The tangent at $(x_1,y_1)$ to the second circle is $S_1=0,$
$x{x}_1+y{y}_1-4(x+x_1)+3(y+y_1)+20=0$
It passes through $(1,-2)$
$\therefore -3x_1+y_1+10=0$ .........$\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$ we get
$x_1=3,y_1=-1$