Answer:
Therefore ,the equation for the circle with a diameter whose end points are (-3,4) and (2,-2) is
[tex]x^{2}+y^{2}+x-2y-14=0[/tex]
Step-by-step explanation:
Given:
End point of Diameter be
point A( x₁ , y₁) ≡ ( -3 , 4 )
point B( x₂ , y₂) ≡ ( 2 , -2 )
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 we get
[tex](x-(-3))(x-2)+(y-4)(y-(-2))=0\\(x+3)(x-2)+(y-4)(y+2))=0[/tex]
Applying Distributive Property A(B+C)=AB+AC we get
[tex]x^{2}+x-6+y^{2}-2y-8=0\\x^{2}+y^{2}+x-2y-14=0[/tex] ..As Required
Therefore ,the equation for the circle with a diameter whose end points are (-3,4) and (2,-2) is
[tex]x^{2}+y^{2}+x-2y-14=0[/tex]
