The least and the greatest distances of the p | vedclass

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Question

(A) Given the circle equation: $x^{2} + y^{2} - 4x - 2y - 20 = 0$. Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -2, f = -1, c = -20$

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$5, 15$

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(A) Given the circle equation: $x^{2} + y^{2} - 4x - 2y - 20 = 0$. Comparing with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -2, f = -1, c = -20$. Centre of the circle is $(-g, -f) = (2, 1)$. Radius $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(-2)^{2} + (-1)^{2} - (-20)} = \sqrt{4 + 1 + 20} = \sqrt{25} = 5$. Let $P$ be the point $(10, 7)$ and $C$ be the centre $(2, 1)$. The distance $d$ between $P$ and $C$ is $\sqrt{(10 - 2)^{2} + (7 - 1)^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$. The least distance of the point from the circle is $d - r = 10 - 5 = 5$. The greatest distance of the point from the circle is $d + r = 10 + 5 = 15$. Thus,the least and greatest distances are $5$ and $15$ respectively.

The least and the greatest distances of the point $(10, 7)$ from the circle $x^{2} + y^{2} - 4x - 2y - 20 = 0$ are

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