Question

Find the equation for the circle with a diameter whose end points are (-3,4) and (2,-2)

Answer 1

Answer:

Therefore ,the equation for the circle with a diameter whose end points are (-3,4) and (2,-2) is

$x^{2}+y^{2}+x-2y-14=0$

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

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

Substituting the end point we get

$(x-(-3))(x-2)+(y-4)(y-(-2))=0\\(x+3)(x-2)+(y-4)(y+2))=0$

Applying Distributive Property A(B+C)=AB+AC we get

$x^{2}+x-6+y^{2}-2y-8=0\\x^{2}+y^{2}+x-2y-14=0$     ..As Required

Therefore ,the equation for the circle with a diameter whose end points are (-3,4) and (2,-2) is

$x^{2}+y^{2}+x-2y-14=0$