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# The equation of the circle whose center is at (3, 7) and whose radius is 6 is

**Solution:**

The equation of the circle is

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

Where (h, k) is the center

r is the radius

It is given that

(h, k) = (3, 7)

r = 6

Substituting these values

(x - 3)^{2} + (y - 7)^{2} = 6^{2}

Expanding using the formula

(a - b)^{2} = a^{2} + b^{2} - 2ab

x^{2} + 9 - 6x + y^{2} + 49 - 14y = 36

By further calculation

x^{2} + y^{2} - 6x - 14y + 9 + 49 = 36

x^{2} + y^{2} - 6x - 14y + 22 = 0

Therefore, the equation of the circle is x^{2} + y^{2} - 6x - 14y + 22 = 0.

## The equation of the circle whose center is at (3, 7) and whose radius is 6 is

**Summary:**

The equation of the circle whose center is at (3, 7) and whose radius is 6 is x^{2} + y^{2} - 6x - 14y + 22 = 0.

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