The maximum distance of the point P(10, 7) fr | vedclass

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(B) The equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. The center $C$ is $(2, 1)$ and the radius $r$ is $\sqrt{2^2 + 1^2 - (-20)} = \sqrt{4

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) The equation of the circle is $x^2 + y^2 - 4x - 2y - 20 = 0$. The center $C$ is $(2, 1)$ and the radius $r$ is $\sqrt{2^2 + 1^2 - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$.The distance from point $P(10, 7)$ to the center $C(2, 1)$ is $PC = \sqrt{(10 - 2)^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$.Since $PC > r$ $(10 > 5)$,the point $P$ lies outside the circle.The maximum distance from point $P$ to the circle is given by $PC + r = 10 + 5 = 15$.

The maximum distance of the point $P(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ is:

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