Question

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

• A. $(-2,1)$
• B. $(3,-1)$
• C. $(-1,-3)$
• D. $(5,0)$

Answer 1

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

Equation of first circle

$x^2+y^2=5$

Equation of tangent of point (1,-2)

$x{x}_1+y{y}_1=5$

$x\left(1\right)+y(-2)=5$

$x-2y=5$ - (1)

Now this equation is a tangent to the other circle too.

$x^2+y^2-8x+6y+20=0$

Replace the term $x$ from equation (1)

$(5+2y{)}^2+y^2-8(5+2y)+6y+20=0$

$\text{On solving;}5y^2+10y+5=0$

$(y+1{)}^2=0$

$y=-1$

$\text{So;}x=5+2(-1)=3$

Hence the point of contact is $(3,-1)$