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Mathematics Question on Application of derivatives

The least and the greatest distances of the point (10,7)(10, 7) from the circle x2+y2−4x−2y−20=0x^2 + y^2 - 4x - 2y - 20 = 0 are

A

10,5

B

5,20

C

44911

D

44696

Answer

44696

Explanation

Solution

Given, x2+y2−4x−2y−20=0x^{2} + y^{2}- 4x - 2y - 20 = 0
Here, g=−2,f=−1,c=−20g =- 2, f =- 1, c = - 20
Centre = (−g,−f)=(2,1)\left(- g, - f\right) = \left(2, 1\right)
Radius
=g2+f2−c=(−2)2+(−1)2−(−20)= \sqrt{g^{2}+f^{2}-c} = \sqrt{\left(-2\right)^{2}+\left(-1\right)^{2}-\left(-20\right)}
=4+1+20= \sqrt{4+1+20}
=25=5= \sqrt{25} = 5

i.e., The distance between the points (2, 1) and
(10,7)=(10−2)2+(7−1)2\left(10, 7\right) = \sqrt{\left(10-2\right)^{2}+\left(7-1\right)^{2}}
=64+36= \sqrt{64+36}
=100=10= \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.