Question

lf a line through $P(-2,3)$ meets the circle $x^2+y^2-4x+2y+k=0$ $at$ $A$ and $B$ such that $PA.PB=31$ then the radius of the circle is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

By property, $PA.PB=PD.PE=P{T}^2$

So, P (-2,3) meets circle $x^2+y^2-4x+2y+K=0$
At ACB $\therefore PA.PB=(PT{)}^2$

Length of tangent to a circle from any point $P(x_1,y_1)$ is

$L=\sqrt{S_1}=\sqrt{x_1^2+y_1^2-4x_1+2y_1+k}$

$=\sqrt{4+9+8+6+K}$

$=\sqrt{27+K}$

Given $31=PA.PB=(PT{)}^2$

$\Rightarrow (27+K)=31$

$\Rightarrow K=4$

Radius of circle $=\sqrt{g^2+f^2-K}$

$=\sqrt{2^2+1^2-4}$

$=\sqrt{1}=1.$