Question

Consider the circle $x^2+y^2-4x-2y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment PA meets the circle in Q with $PQ=5$, then $c=$

• A. $-15$
• B. $20$
• C. $30$
• D. $-20$

Answer 1

The correct option is D $-20$

Given equation is

${(x-2)}^2+{(y-1)}^2=5-c$

Now

$PA=\sqrt{64+36}=10$

$PQ=5$ (Given)

$\therefore$ $AQ=10-5=5$

$\therefore$ Radius $=5$ unit

So, $5-c=25$

$\Rightarrow$ $c=-20$