Question

The shortest distance from the point (2, –7) to the circle x² + y² – 14x – 10y – 151 = 0 is equal to 5.

Options

• True

• False

Answer 1

This statement is False.

Explanation:

Given equation of circle is x² + y² – 14x – 10y – 151 = 0

Shortest distance = distance between the point (2, – 7)

And the centre – radius of the circle

Centre of the given circle is

2g = – 14 ⇒ g = – 7

2f = – 10 ⇒ f = – 5

∴ Centre = (– g, – f) = (7, 5)

And r = `sqrt((-7)^2 + (-5)^2 + 151)`

= `sqrt(49 + 25 + 151)`

= `sqrt(225)`

= 15

∴ Shortest distance = `sqrt((7 - 2)^2 + (5 + 7)^2) - 15` 

= `sqrt(25 + 144) - 15`

= 13 – 15

= |– 2|

= 2