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# Find the equation for a circle given center at (-1,-2) & passes through the point (15,-2)?

**Solution:**

If the a point P(x, y) lies on the circle with the centre at (h, k) then the general equation of the circle will be written as

\(\sqrt{(x-h)^{2} + (x-k)^{2}} = a\)

where a is the radius

The radius of the circle is calculated using the distance formula as:

\(\sqrt{(15-(-1)^{2} + (-2-(-2))^{2})}\) = \(\sqrt{256 +0}\) = 16

The general equation of the circle can be therefore written as :

\(\sqrt{(x-(-1)^{2} + (y-(-2))^{2}} = 16\)

Squaring both the sides and solving we get the general equation of the circle asl

x^{2} + 1 + 2x + y^{2} + 4 + 4y = 256

x^{2} + y^{2} + 2x + 4y - 251 = 0

## Find the equation for a circle given center at (-1,-2) & passes through the point (15,-2)?

**Summary:**

The equation of the circle with center (-1,-2) and containing the point (15,-2) is x^{2} + y^{2} + 2x + 4y - 251 = 0

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