Question

The length of tangent from an external point to the circle $x^2+y^2-2x-2y-7=0$ is 4 units. What is the distance of this point to the farthest point in the circle?

Answer 1

Given circle $\equiv x^2+y^2-2x-2y-7=0$

In standard form,

$(x-1{)}^2+(y-1{)}^2=3^2$

$\therefore$ Radius $\equiv$ 3 units

Since $△$PTO is right angled

$P{O}^2=P{T}^2+T{O}^2$

$\therefore PO=\sqrt{3^2+4^2}=5$

Farthest point will be the one which we get on extending the line PO to intersect the circle.

$\therefore PF=PO+OF=PO+radius$

$=5+3=8units$