# Write the equation of the circle with center (3, 2) and (-6, -4) a point on the circle.

**Solution:**

The standard form of the equation of a circle is given by

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

Where, a and b are the coordinates of the centre

r is the radius

Given, centre = (3, 2) and (x,y) = (-6 , -4)

Substituting the values in standard form of equation we get,

(-6 - 3)^{2} + (-4 - 2)^{2} = r^{2}

(-9)^{2} + (-6)^{2} = r^{2}

r^{2} = 81 + 36

r = √117

Thus, the radius of the circle is √117 units.

The equation of a circle can be written as

(x - 3)^{2} + (y - 2)^{2} = (√117)^{2}

(x - 3)^{2} + (y - 2)^{2} = 117

Therefore, the equation of a circle is (x - 3)^{2} + (y - 2)^{2} = 117.

## Write the equation of the circle with center (3, 2) and (-6, -4) a point on the circle.

**Summary:**

The equation of the circle with center (3, 2) and (-6, -4) a point on the circle is (x - 3)^{2} + (y - 2)^{2} = 117.

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