# Find the equation of the circle whose center = ( 5, 6) and radius = 3?

**Solution:**

Given: Center of the circle (5, 6) and radius = 3

We know that the equation of a circle whose center is (h,k) and radius is r can be given as

(x - h)^{2} + (y - k)^{2} = r^{2} --- (1)

Substitute the values h = 5 , k = 6 and r = 3 in the equation(1)

(x - 5)^{2} + (y - 6)^{2} = 3^{2}

x^{2} - 10x + 25 + y^{2} - 12y + 36 = 9

x^{2} + y^{2} - 10x - 12y + 52 = 0

Thus the equation of the circle with given data is x^{2} + y^{2} - 10x - 12y + 52 = 0

## Find the equation of the circle whose center = ( 5, 6) and radius = 3?

**Summary:**

The equation of the circle with centre (5, 6) and radius = 3 is x^{2} +y^{2} - 10x - 12y + 52 = 0.

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