Question

A circle with centre (-3, 1) passes through the point (3, 1).find the equation

Answer 1

The general formula for a circle with center (a,b) and radius r is

$(x-a)^2 + (y-b)^2 = r^2$

If we know the center and a point on the circle, the squared radius is given by substituting in the point for x and y.

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

Answer:

$(x+3)^2 + (y-1)^2 = 36$

Answer 2

To write the equation of a circle, you need its radius and its center. The center is given, and you can easily derive the radius: it is the distance between the center and any point in the circumference.

In your case, the center is $(-3,1)$, and the point on the circumference is $(3,1)$. These points are 6 units apart, so the radius is 6.

Now, knowing the center $(k,h)$ and the radius $r$, the equation is

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

Plugging your values, you have

$(x-(-3))^2+(y-1)^2 = 6^2 \iff (x+3)^2+(y-1)^2 = 36$