Question

If the tangent to the circle $x^2+y^2=5$ at the point $(1,-2)$ also touches the circle $x^2+y^2-8x+6y+20=0$, then its point of contact is

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

Answer 1

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

Equation of tangent to circle at $x^2+y^2=5$ at $(1,-2)$ center $(0,0)$

Slope of line from center to point $(1,-2)$

$m=\frac{-2-0}{1-0}=-2$

Slope of tangent $=\frac{1}{2}$

Equation of tangent

$(y+2)=\frac{1}{2}(x-1)2y=x-5$

This tangent also touches the circle $x^2+y^2-8x+6y+20=0$

Intersection point is

$x^2+{\left(\frac{x-5}{2}\right)}^2-8x+\frac{6}{2}(x-5)+20=04x^2+{(x-5)}^2-32x+12(x-5)+80=05x^2-30x+45=0x^2-6x+9=0\Rightarrow {(x-3)}^2=0\Rightarrow x=3$

Putting this value of $x$ in $2y=x-5y=-1$

Point of contact $(3,-1)$