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# 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} = r^{2}

(x - 2)^{2 }+ (y + 1)^{2} = (√26)^{2}

Expanding using the algebraic identity

(a + b)^{2} = a^{2} + b^{2} + 2ab

(a - b)^{2} = a^{2} + b^{2} - 2ab

x^{2} + 4 - 4x + y^{2} + 1 +2y = 26

So we get

x^{2} - 4x + 2y + y^{2} + 5 = 26

Therefore, the equation of the circle is x^{2 }- 4x + 2y + y^{2} - 19 = 0.

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

**Summary:**

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

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