Find the maximum distance of the point K(10, | vedclass

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Question

(C) The equation of the given circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Completing the square,we get $(x - 2)^2 + (y - 1)^2 = 25$. This circle has cen

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The equation of the given circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. Completing the square,we get $(x - 2)^2 + (y - 1)^2 = 25$. This circle has center $C(2, 1)$ and radius $r = \sqrt{25} = 5$. The distance between the point $K(10, 7)$ and the center $C(2, 1)$ is $CK = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$. The maximum distance of the point $K$ from the circle is given by $CK + r = 10 + 5 = 15 	ext{ units}$. Thus,option $C$ is correct.

Find the maximum distance of the point $K(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$.

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