# What is the equation of a circle with radius 4 that passes through (1, 4) and (2, 6)

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### 1 Answer

The equation of a circle has to be determined that has a radius 4 and passes through the points (1, 4) and (2, 6)

Let the equation of the circle be (x - a)^2 + (y - b)^2 = 16

As it passes through (1, 4)

(1 - a)^2 + (4 - b)^2 = 16

=> 1 + a^2 - 2a + 16 + b^2 - 8b = 16

=> a^2 + b^2 - 2a - 8b + 1 = 0 ...(1)

And as it passes through (2, 6)

(2 - a)^2 + (6 - b)^2 = 16

=> 4 + a^2 - 4a + 36 + b^2 - 12b = 16

=> a^2 + b^2 - 4a - 12b + 24 = 0 ...(2)

(1) - (2)

=> 2a + 4b - 23 = 0

=> a = `(23 - 4b)/2`

Substitute in (2)

`(23 - 4b)^2/4 + b^2 - 2(23 - 4b) - 12b + 24 = 0`

Solving the equation gives the roots

b1 = `-(sqrt(295)-50)/10` and b2 = `(sqrt(295)+50)/10`

a1 = `(2*sqrt(295)+15)/10` and a2 = `-(2*sqrt(295)-15)/10`

**The equation of the required circle is:**

`(x - (2*sqrt(295)+15)/10)^2 + (y + (sqrt(295)-50)/10)^2 = 16` and `(x + (2*sqrt(295)-15)/10)^2 + (y -(sqrt(295)+50)/10)^2 = 16`