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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 $(1,4), (6,5)$.

Elaboration: If you draw a line through two points on a circle, find the mid-point of the line and then draw a perpendicular line through the mid-point, then this line will pass through the centre of the circle. If you take the intersection of this line and the line on which the centre lies, then you will find the circle centre.

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  • $\begingroup$ I don't understand please elaborate. $\endgroup$ – Nick Dec 9 '13 at 15:25
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    $\begingroup$ I added some elaboration. $\endgroup$ – copper.hat Dec 9 '13 at 16:00

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