Answer:
The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is
[tex]x^{2} +y^{2}+4x-1=0[/tex]
Step-by-step explanation:
Explanation:-
Step 1:-
The equation of the circle having center and radius is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
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
[tex](\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )[/tex]
[tex](-2,0)[/tex]
therefore center (h,k) = (-2,0)
Step 2:-
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
[tex]\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}[/tex]
[tex]r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}[/tex]
[tex]r=\sqrt{5}[/tex]
Step 3:-
center (h,k) = (-2,0) and
radius [tex]r=\sqrt{5}[/tex]
The standard form of circle equation
[tex](x-h)^2+(y-k)^2=r^2[/tex]
[tex](x-(-2))^2+(y-0)^2=\sqrt{5} ^2[/tex]
on simplification is
[tex]x^{2} +y^{2}+4 x-1=0[/tex]