# How do you find the equation of a circle passing through (2,3), (3,4) and (-1,2)?

Sep 15, 2016

$\text{ Eqn. of the circle : } {x}^{2} + {y}^{2} + 2 x - 14 y + 25 = 0$.

#### Explanation:

There are more than one Methods to solve this problem.

I select the following one, which is the easiest in my opinion.

Let, the eqn. of the Circle be

$S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 , . . \left(\star\right)$, known as the General

Eqn. of Circle.

$\text{ Pt. } \left(2 , 3\right) \in S \Rightarrow 4 + 9 + 4 g + 6 f + c = 0. \ldots \ldots \left(1\right)$

$\text{ Pt. } \left(3 , 4\right) \in S \Rightarrow 9 + 16 + 6 g + 8 f + c = 0. \ldots . . \left(2\right)$

$\text{ Pt. } \left(- 1 , 2\right) \in S \Rightarrow 1 + 4 - 2 g + 4 f + c = 0. \ldots \ldots . . \left(3\right)$

$\left(2\right) - \left(1\right) \Rightarrow 12 + 2 g + 2 f = 0. \ldots \ldots \ldots \ldots \ldots . . \left(4\right)$

$\left(3\right) - \left(1\right) \Rightarrow - 8 - 6 g - 2 f + 0. \ldots \ldots \ldots \ldots \ldots . . \left(5\right)$

$\left(4\right) + \left(5\right) \Rightarrow 4 - 4 g = 0 \Rightarrow g = 1. \text{ Now, } \left(4\right) \Rightarrow f = - 7.$

 g=1, f=-7, &, (3) rArr c=25.

Using all these consts. in $\left(\star\right)$, we have, the eqn. of circle,

$S : {x}^{2} + {y}^{2} + 2 x - 14 y + 25 = 0.$

Enjoy Maths.!