Respuesta :

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]

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