Question

Minimum distance of the circle x^2+y^2-4x-2y-20=0 from point (10,7)

Answer 1

hmmmmm..

.............sorryyyy

Answer 2

Answer:

The circle is

 x² + y² - 4x - 2y - 20 = 0

⇒ (x - 2)² + (y - 1)² = 20 + 4 + 1

⇒ (x - 2)² + (y - 1)² = 5²,

whose centre is at (2, 1) and radius is 5 units.

Now, the distance between the points (2, 1) and (10, 7) is

= 10 units

So, the radius is less than the distance between the points and thus P lies outside the circle.

Thus, the greatest distance is

 = radius + the distance between the centre of the circle and P (10, 7)

 = (5 + 10) units

 = 15 units.