How do you write an equation for a circle whose diameter has endpoints of (-2, 4) and (4, 12)?

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Answer 1 · 2018-08-14T22:27:48

The eqn . of circle :
$x^2+y^2-2x-16y+40=0$

Let , $A(-2,4) and B(4,12)$ be the endpoints of diameter $bar(AB)$ and $C$ be the center of given circle.

Let $r$ be the radius of circle.

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Here ,

Center $C$ is the midpoint of diameter $bar(AB)$

So ,

$C((-2+4)/2,(4+12)/2)-=(1,8)$

Now ,,

$"Radius r="CA=CB$

$:.r=CB=sqrt((4-1)^2+(12-8)^2)=sqrt(9+16)=5$

We know that,

$"The eqn. of circle with center C(h,k) and radius r "$ is

$(x-h)^2+(y-k)^2=r^2$

$:.(x-1)^2+(y-8)^2=5^2$

$:.x^2-2x+1+y^2-16y+64=25$

$:.x^2+y^2-2x-16y+40=0$

Answer 2 · 2018-08-14T22:43:03

The eqn. of circle is :
$:.x^2+y^2-2x-16y+40=0$

We know that ,

$"If "A(x_1,y_1) and B(x_2,y_2)"are the endpoints of the "$

$"diameter of the circle then the eqn. of circle is :"$

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

We have diameter $bar(AB)$ ,with $A(-2,4) and B(4,12).$

So, the eqn. of circle is :

$(x-(-2))(x-4)+(y-4)(y-12)=0$

$(x+2)(x-4)+(y-4)(y-12)=0$

$:.x^2-4x+2x-8+y^2-12y-4y+48=0$

$:.x^2+y^2-2x-16y+40=0$