# Finding the equation of a circle from given points on it and line on which the centre lies.

What are some effective ways to find the equation of a circle when you are given points lying on the circle and the equation for the line on which the centre of the circle lies.

Here is an example of such a question:

Find the equation of a circle passing through the points $(4,1)$ and $(6,5)$ and whose centre lies on the line $4x + y = 16$

I have figured out one way to get the solution.

Let the $(h,k)$ be center of the circle, Since the line $4x + y = 16$ also includes the center $(h,k)$, then the following must be true,$$4h + k = 16\quad\quad\quad\text{(1)}$$

Let $r$ be the radius and $(x,y)$ be any point on the circle. Then, the equation satisfied by all points of the circle is $$(x-h)^2 + (y-k)^2 = r^2$$

Using this equation, $$(4-h)^2 - (1-k)^2 = r^2\quad\wedge\quad(6-h)^2 + (5-k)^2 = r^2$$ On equating and simplifying ,$$h + 2k = 11 \quad\quad\quad\text{(2)}$$

On solving the linear equations $\text{(1) and (2)}$, $$h = 3,\space k= 4 \quad\Rightarrow\quad \text{The centre is } (3,4)$$

Sticking that into an equation satisfied by a given point, $$r^2 = (6-3)^2 + (5-4)^2 \quad\Rightarrow\quad r = \sqrt 10$$

I could have also used the distance formula ($d = \sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}$) to find the radius.

But I want another approach. It feels too limiting just have only one or two ways to answer a question. I need have new perspectives. Please enlighten me with a different way to do this.

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Well, it amounts to the same thing, but take the intersection of the line with the bisector of the line joining $(1,4), (6,5)$.