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

10, 5

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15, 20

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12, 16

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5, 15

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Solution

The correct option is D $5, 15$
The equation of the given circle is $x^{2} + y^{2} - 4 x - 2 y - 20 = 0$ and when substituted $x = 10$ & $y = 7$ , its value becomes $100 + 49 - 40 - 14 - 20 = 75$ which is greater than zero.
Thus, the point $(10, 7)$ lies outside the circle.
Its distance from the centre of the circle $(2, 1)$ is $\sqrt{(10 - 2)^{2} + (7 - 1)^{2}} = \sqrt{100} = 10$ unit.
The minimum distance from the circle therefore becomes $10 - 5$ (i.e. radius) $= 5$ and the maximum distance becomes $10 + 5 = 15$ unit.

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Similar questions

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

(10, 7)

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

(10, 7)

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

P (10, 7)

x^{2} + y^{2} - 4 x - 2 y - 20 = 0

P (10, 7)

C : x^{2} + y^{2} - 4 x - 2 y - 20 = 0

l, m

C

{(9 l + 2 m)}^{2}

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