Find the equation of a circle that passes through (7, -1) and has a center of (-2, 4)?

Solution:

Circle is a set of all points in the plane whose distance from the fixed point is equal.

The fixed point is called the center of the circle and the constant distance is called the radius of the circle. The circle's equations can be identified in two ways. One is the equation of the circle with the center at the origin and radius ‘a’ the equation is x² + y² = a².

We have the equation of a circle centered at (h, k) and Radius ‘r’ as (x - h)² + (y - k)² = r²

Here, center (h, k) = (-2, 4) ⇒ h = - 2 and k = 4

The radius of the circle is the distance between the points (7, -1) and (-2, 4)

We have distance between the points (x₁ , y₁) and (x₂ , y₂) is √{(x₂ - x₁)² + (y₂ - y₁)²}

∴ Radius = r = √{(7 - (-2))² + (-1 - 4)² } = √(81 + 25) = √106

Thus the equation of the circle is

[x - (-2)]² + (y - 4)² = {√106}²

⇒ x² + y² + 4x - 8y + 4 + 16 = 106

⇒ x² + y² + 4x - 8y - 86 = 0

Find the equation of a circle that passes through (7, -1) and has a center of (-2, 4)?

Summary:

The equation of a circle that passes through (7, -1) and has a center of (-2, 4) is x² + y² + 4x - 8y - 86 = 0