The greatest distance of the point $P (9, 7),$ from the circle $x^{2} + y^{2} - 2 x - 2 y - 23 = 0,$ is units

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Solution

Since,
$S_{1} = 9^{2} + 7^{2} - (2 \times 9) - (2 \times 7) - 23 \Rightarrow S_{1} = 75 > 0$
So, $P$ lies outside the circle.
Coordinates of the centre of the circle is $C (1, 1)$ and radius is $r = 5$ units
Hence, the greatest distance of point $P$ from the circle
$= P C + r = \sqrt{8^{2} + 6^{2}} + 5 = 15 units$

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Position of a Point with Respect to Circle

Standard XIII Mathematics

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The greatest distance of the point P(9,7), from the circle x2+y2−2x−2y−23=0, is units

Since, S1=92+72−(2×9)−(2×7)−23⇒S1=75>0 So, P lies outside the circle. Coordinates of the centre of the circle is C(1,1) and radius is r=5 units Hence, the greatest distance of point P from the circle =PC+r=√82+62+5=15 units

The greatest distance of the point $P (9, 7),$ from the circle $x^{2} + y^{2} - 2 x - 2 y - 23 = 0,$ is units