Question

Find the equation of the circle having $(6,-3)$ and $(2,-1)$ as the endpoints of its diameter.

Answer 1

We know that the center of a circle is the mid point, of end points of its diameter.
$(6,-3)$ and $(2,-1)$ are given as endpoints of diameter.
If we name O as the center of the circle then
From the mid-point formula :
$\text{O}(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\text{Let,}{x}_1=6,x_2=2y_1=-3,y_2=-1$
So $\text{O}(x,y)=(\frac{6+2}{2},\frac{-3-1}{2})$
$\text{O}(x,y)=(4,-2$)

Also we know that the diameter is twice than radius so we can find the radius also using distance formula.
We know, the distance between $(x_1,y_1)and(x_2,y_2)$
= $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Diameter = $\sqrt{(2-6{)}^2+(-1+3{)}^2}$
= $\sqrt{(-4{)}^2+(2{)}^2}$
= $\sqrt{20}$
= 2$\sqrt{5}$
Radius = $\sqrt{5}$
Now we have the coordinates of center and radius.
Equation of a circle with centre $(h,k)$ and radius r is given by
$(x-h{)}^2+(y-k{)}^2=r^2$

Circle = $(x-4{)}^2+(y+2{)}^2=5$
= $x^2+y^2-8x+4y+16+4=5$
Equation : $x^2+y^2-8x+4y+15=0$