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# How do you write an equation for a circle whose diameter has endpoints of $\left( -2,4 \right)$ and $\left( 4,12 \right)$?

Last updated date: 29th Feb 2024
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Hint: We should know the standard form equations to solve this question. First we have to know the formula for the circle whose diameter has endpoints. Then we have to substitute given endpoints in the formula and simplify it to get the equation of the circle.

The standard equation of the circle with center $\left( h,k \right)$ and radius $r$ is
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$.
The equation of the circle diameter with endpoints $\left( {{x}_{1}},{{x}_{2}} \right)$and $\left( {{y}_{1}},{{y}_{2}} \right)$ is
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$.
We will use the above formula to solve our problem.
Given endpoints are $\left( -2,4 \right)$ and $\left( 4,12 \right)$
We already discussed the equation of circle whose diameter with endpoints $\left( {{x}_{1}},{{x}_{2}} \right)$and $\left( {{y}_{1}},{{y}_{2}} \right)$ is
given by
$\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0$
Now we have the circle whose diameter has endpoints $\left( -2,4 \right)$ and $\left( 4,12 \right)$.
So we substitute these formulas in the equation we have.
After substitution we will get the equation as
$\Rightarrow \left( x-\left( -2 \right) \right)\left( x-4 \right)+\left( y-4 \right)\left( y-12 \right)=0$
By simplifying we will get
$\Rightarrow \left( x+2 \right)\left( x-4 \right)+\left( y-4 \right)\left( y-12 \right)=0$
Now we have to multiply the terms with each other.
By multiplying first two terms we will get
$\Rightarrow {{x}^{2}}+2x-4x-8+\left( y-4 \right)\left( y-12 \right)=0$
Now we have to multiply the next two terms.
$\Rightarrow {{x}^{2}}+2x-4x-8+{{y}^{2}}-12y-4y+48=0$
Now we have to simplify the equation accordingly to arrive at the solution.
Now we have to add or subtract the like terms. We will get
$\Rightarrow {{x}^{2}}-2x-8+{{y}^{2}}-16y+48=0$
By further simplifying we will get
$\Rightarrow {{x}^{2}}-2x+{{y}^{2}}-16y+40=0$
Now we have to rearrange the terms accordingly. We will get
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-16y+40=0$
So the equation of the circle whose diameter has endpoints $\left( -2,4 \right)$ and $\left( 4,12 \right)$ is
${{x}^{2}}+{{y}^{2}}-2x-16y+40=0$.

Note:
We can also do this in another way. First we have to find the center of the diameter and then we have to find the radius of the circle. After that we have to substitute in the standard equation of circle. We can do it in either way. But the above said method is the simplest one.