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 

x² + 4 - 4x + y² + 16 - 8y = 25

x² + y² - 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  x² + y² - 4x - 8y - 5 = 0