Answer:
The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is
![x^{2} +y^{2}+4x-1=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%2B4x-1%3D0)
Step-by-step explanation:
<u>Explanation:</u>-
<u>Step 1:</u>-
The equation of the circle having center and radius is
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
here center is (h,k) and radius is r
Given diameter whose end points are (-1,-2) and (-3,2)
The diameter of the circle is passing through the center of the circle
so center of the circle = midpoint of two end points
![(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )](https://tex.z-dn.net/?f=%28%5Cfrac%7B-1%20%2B%28-3%29%20%7D%7B2%7D%20%2C%5Cfrac%7B-2%2B2%20%7D%7B2%7D%20%20%29)
![(-2,0)](https://tex.z-dn.net/?f=%28-2%2C0%29)
therefore center (h,k) = (-2,0)
<u>Step 2:-</u>
we have to find the radius of the circle
the radius of the circle = the distance from center to the one end point
i.e., C P = r
Given one end point is P(-3,2) and center C(-2,0)
The distance formula of two points are
![\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}](https://tex.z-dn.net/?f=%5Csqrt%7B%28x_%7B2%7D-x_%7B1%7D%20%29%20%5E%7B2%7D%2B%20%28y_%7B2%7D-y_%7B1%7D%20%29%20%5E%7B2%7D%7D)
![r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%7B%28-3%29-%28-1%29%20%29%20%5E%7B2%7D%2B%20%282-%28-2%29%29%20%5E%7B2%7D%7D)
![r=\sqrt{5}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B5%7D)
<u>Step 3</u>:-
center (h,k) = (-2,0) and
radius ![r=\sqrt{5}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B5%7D)
The standard form of circle equation
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
![(x-(-2))^2+(y-0)^2=\sqrt{5} ^2](https://tex.z-dn.net/?f=%28x-%28-2%29%29%5E2%2B%28y-0%29%5E2%3D%5Csqrt%7B5%7D%20%5E2)
on simplification is
![x^{2} +y^{2}+4 x-1=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2By%5E%7B2%7D%2B4%20x-1%3D0)