Question

The tangent to the circle $x^2+y^2-4x+2y+k=0$ at $(1,1)$ is $x-2y+1=0$ then $k=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

Eqn of circle $\Rightarrow x^2+y^2-4x+2j+k=0$

general form $\Rightarrow x^2+y^2+bx+cy+k=0$

centre of circle $(\frac{-b}{2},\frac{-c}{2})=(2,-1)$

$radius=\sqrt{(2+1{)}^2+(-1-1{)}^2}=\sqrt{1+4}=\sqrt{5}$

also radius$=\sqrt{{\left(\frac{b}{2}\right)}^2+{\left(\frac{c}{2}\right)}^2-k}=\sqrt{5}$

$={\left(\frac{-4}{2}\right)}^2+{\left(\frac{2}{2}\right)}^2-k=5$

$4+1-k=5$

$k=0$