Question

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

Answer 1

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`