How to find the equation of a circle given the center(2, -1) and a point it passes through (3, 4)?

Solution:

The centre of the circle is given as (h, k) = (2, -1)

Since, the circle passes through the point (3,4), the radius (r) of the circle is the distance between the point (2, -1) and (3,4).

r = √(3 - 2)² + (4 + 1)²

r = √1 + 25 = √26

The equation of the circle is (x - h)²+ (y - k)² = r²

(x - 2)²+ (y + 1)² = (√26)²

Expanding using the algebraic identity

(a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

x² + 4 - 4x + y² + 1 +2y = 26

So we get

x² - 4x + 2y + y² + 5 = 26

Therefore, the equation of the circle is x²- 4x + 2y + y² - 19 = 0.

Summary:

The equation of a circle given the center(2, -1) and a point it passes through (3, 4) is x² - 4x + 2y + y²- 19 = 0.

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