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vlabodo [156]
3 years ago
15

Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
5 0

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

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

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}  )

    (-2,0)

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}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

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