Question

Write the equation of a circle for which the endpoints of a diameter are (-2,-2) and (4,-10)

Answer 1

Answer:

Therefore the required Equation of Circle

$x^{2}+y^{2}-2x+12y+12=0$

Step-by-step explanation:

Given:

End point of Diameter be

point A( x₁ , y₁) ≡ ( -2 ,-2 )

point B( x₂ , y₂) ≡ ( 4 , -10 )

To Find:

Equation of a circle =?

Solution:

When end points of the Diameter are A( x₁ , y₁) , B( x₂ , y₂). then the Equation of Circle is given as

$(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$

Substituting the end point are

$(x-(-2))(x-4)+(y-(-2))(y-(-10))=0\\(x+2))(x-4)+(y+2))(y+10))=0$

Applying Distributive Property we get

$x^{2} -2x-8+y^{2}+12y+20 =0\\\\x^{2}+y^{2}-2x+12y+12=0$

Therefore the required Equation of Circle

$x^{2}+y^{2}-2x+12y+12=0$