Question

How do you find the equation of the circle given diameter at (4, -3) and (-2, 5)?

Answer 1

`(x-1)^2+(y-1)^2=25`

Explanation:

If the diameter is `color(brown)(""(4,-3))` to `color(orange)(""(-2,5))`
then center is at
`color(white)("XXX")((color(brown)(4)+(color(orange)(-2)))/2,(color(brown)(-3)+color(orange)(5))/2)=(color(red)(1),color(blue)(1))`
and
the radius is
`color(white)("XXX")sqrt((color(brown)(4)-color(red)(1))^2+(color(brown)(-3)-color(blue)(1))^2)=sqrt(9+16) =color(green)(5)`

The general equation of a circle
with center `(color(red)(a),color(blue)(b))` and radius `color(green)(r)` is
`color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))^2=color(green)(r)^2`

In this case we have
`color(white)("XXX")(x-color(red)(1))^2+(y-color(blue)(1))^2=color(green)(5)^2`

Answer 2

`x^2+y^2-2x-2y-23=0`,

Explanation:

Formula for the eqn. of a circle having `(x_1,y_1), (x_2,y_2)` as diametrically opposite points (pts.) : `(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.`

Accordingly, the reqd. eqn. is `(x-4)(x+2)+(y+3)(y-5)=0`,
i.e., `x^2-4x+2x-8+y^2+3y-5y-15=0`,
or, `x^2+y^2-2x-2y-23=0`,

Aliter :-

Extremities of a diameter are `(4,-3), (-2,5)`

Centre, being midpoint of diameter, we get the centre `((4-2)/2,(-3+5)/2)=(1,1).`

Radius = Distance between the Centre & any end point of given diameter `= sqrt {(4-1)^2+(-3-1)^2}=5.`

Hence the eqn. `: (x-1)^2+(y-1)^2=5^2,` or, `x^2+y^2-2x-2y-23=0,` as before!