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# Write the equation of the circle with center (-1, -3) and (-7, -5) 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 = (-1, -3) and (x , y) = (-7, -5)

Substituting the values in standard form of equation we get,

(-7 - (-1))^{2} + (-5 - (-3))^{2} = r^{2}

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

r^{2} = 36 + 4

r = √40

Thus, the radius of the circle is √40 units.

The equation of a circle can be written as

(x - (-1))^{2} + (y - (-3))^{2} = (√40)^{2}

(x + 1)^{2} + (y + 3)^{2} = 40

Therefore, the equation of a circle is (x + 1)^{2} + (y + 3)^{2} = 40.

## Write the equation of the circle with center (-1, -3) and (-7, -5) a point on the circle.

**Summary:**

The equation of the circle with center (-1, -3) and (-7, -5) a point on the circle is (x + 1)^{2} + (y + 3)^{2} = 40.

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