Answer:
Therefore the required Equation of Circle
[tex]x^{2}+y^{2}-2x+12y+12=0[/tex]
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
[tex](x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0[/tex]
Substituting the end point are
[tex](x-(-2))(x-4)+(y-(-2))(y-(-10))=0\\(x+2))(x-4)+(y+2))(y+10))=0\\[/tex]
Applying Distributive Property we get
[tex]x^{2} -2x-8+y^{2}+12y+20 =0\\\\x^{2}+y^{2}-2x+12y+12=0[/tex]
Therefore the required Equation of Circle
[tex]x^{2}+y^{2}-2x+12y+12=0[/tex]
