Question

what is the greatest distance of the point P(10,7) from the circle x2 + y2 - 4x - 2y - 20 = 0 ?

Answer 1

The answer is given below :

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

$=\sqrt{{(10-2)}^{2}+{(7-1)}^{2}}$

$=\sqrt{{8}^{2}+{6}^{2}}$

$=\sqrt{64+36}$

$=\sqrt{100}$

$=\sqrt{(10^{2})}$

    = 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.

Answer 2

Answer:

The answer is given below :

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.