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

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Solution

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

In standard form,

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

$∴$ Radius $\equiv$ 3 units

Since $△$ PTO is right angled

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

$∴ P O = \sqrt{3^{2} + 4^{2}} = 5$

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

$∴ P F = P O + O F = P O + r a d i u s$

$= 5 + 3 = 8 u n i t s$

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