Question:
The least and the greatest distances of the point $(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ are
The least and the greatest distances of the point $(10, 7)$ from the circle $x^2 + y^2 - 4x - 2y - 20 = 0$ are
Updated On: Apr 18, 2024
- 10,5
- 5,20
- 44911
- 44696
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The Correct Option is D
Solution and Explanation
Given, $x^{2} + y^{2}- 4x - 2y - 20 = 0$
Here, $g =- 2, f =- 1, c = - 20$
Centre = $\left(- g, - f\right) = \left(2, 1\right)$
Radius
$= \sqrt{g^{2}+f^{2}-c} = \sqrt{\left(-2\right)^{2}+\left(-1\right)^{2}-\left(-20\right)}$
$= \sqrt{4+1+20}$
$= \sqrt{25} = 5$
i.e., The distance between the points (2, 1) and
$\left(10, 7\right) = \sqrt{\left(10-2\right)^{2}+\left(7-1\right)^{2}}$
$= \sqrt{64+36}$
$= \sqrt{100} = 10$ units
$\therefore$ The least distance of the point (10, 7) from the circle = 10 - 5 = 5 units and the greatest distance of the point (10, 7) from the circle = 10 + 5 = 15 units.
Here, $g =- 2, f =- 1, c = - 20$
Centre = $\left(- g, - f\right) = \left(2, 1\right)$
Radius
$= \sqrt{g^{2}+f^{2}-c} = \sqrt{\left(-2\right)^{2}+\left(-1\right)^{2}-\left(-20\right)}$
$= \sqrt{4+1+20}$
$= \sqrt{25} = 5$
i.e., The distance between the points (2, 1) and
$\left(10, 7\right) = \sqrt{\left(10-2\right)^{2}+\left(7-1\right)^{2}}$
$= \sqrt{64+36}$
$= \sqrt{100} = 10$ units
$\therefore$ The least distance of the point (10, 7) from the circle = 10 - 5 = 5 units and the greatest distance of the point (10, 7) from the circle = 10 + 5 = 15 units.
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