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# Write the equation of the circle with center (2,4) and containing the point (-2,1)?

**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:

\(\sqrt{(2-(-2))^{2} + (4-1)^{2})}\) = \(\sqrt{16 + 9}\) = 5

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

\(\sqrt{(x-2)^{2} + (y-4)^{2}} = 5\)

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

x^{2} + 4 - 4x + y^{2} + 16 - 8y = 25

x^{2} + y^{2} - 4x - 8y - 5 = 0

## Write the equation of the circle with center (2,4) and containing the point (-2,1)?

**Summary:**

The equation of the circle with center (2,4) and containing the point (-2,1) is x^{2} + y^{2} - 4x - 8y - 5 = 0

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