Question

The point (-3, 2) lines on a circle whose equation is (x + 3)2 + ( + 1)2 = r2. Which of

Answer 1

Answer:

3 units

Step-by-step explanation:

Equation of the circle is:

$(x+3)^2 +(y+1)^2=r^2$

$\therefore [x-(-3)] ^2 +[(y-(-1)] ^2=r^2$

Equating it with

$(x-h)^2 +(y-k)^2=r^2$

We find: h = - 3 & k = - 1

Center of the circle (C) = (-3, - 1)

Since, the point (-3, 2) lies on the circle, therefore radius of the circle wii be equal to the distance between the center of the circle (-3, - 1) and the point (-3, 2) which is on the circle.

Therefore,

$r = \sqrt{[-3-(-3)]^2 +[2-(-1)]^2}$

$r = \sqrt{[-3+3]^2 +[2+1]^2}$

$r = \sqrt{[0]^2 +[3]^2}$

$r = \sqrt{0 +9}$

$r = \sqrt{9}$

$r = 3\: units$