Question

How do you use the information provided to write the equation of each circle:Three points on the circle are (-15, -6), (-19,-4), and (-17, -6)?

Answer 1

Equation of only circle is `x^2+y^2+32x+6y+255=0`

Explanation:

As there are three points on the circle are `(-15, -6)`, `(-19,-4)` and `(-17, -6)`, there is only one circle and the center of circle will lie on perpendicular bisector of any two sides.

Let us find the equation of perpendicular bisector of points joining `(-15, -6)` and `(-19,-4)`.

As a point on a perpendicular bisector is equidistant from `(-15, -6)` and `(-19,-4)`, its equation will be

`(x+15)^2+(y+6)^2=(x+19)^2+(y+4)^2`

or `x^2+30x+225+y^2+12y+36=x^2+38x+361+y^2+8y+16`

i.e. `30x-38x+12y-8y=377-261`

or `-8x+4y=116` or `2x-y=-29` ....................(1)

Similarly equation of perpendicular bisector of points joining `(-17, -6)` and `(-19,-4)` is

`(x+17)^2+(y+6)^2=(x+19)^2+(y+4)^2`

or `x^2+34x+289+y^2+12y+36=x^2+38x+361+y^2+8y+16`

i.e. `34x-38x+12y-8y=377-325`

or `-4x+4y=52` or `x-y=-13` ....................(2)

Subtracting equation (2) from (1), we get

`x=-16` and hence putting this in (2), we get `y=-3`

Hence center of the circle is `(-16,-3)` and square of the radius is square of the distance of center from say `(-17,-6)` i.e.

`(-16+17)^2+(-3+6)^2=1+9=10`

Hence equation of circle is

`(x+16)^2+(y+3)^2=10`

or `x^2+32x+256+y^2+6y+9=10`

or `x^2+y^2+32x+6y+255=0`
graph{x^2+y^2+32x+6y+255=0 [-20.645, -10.645, -6.88, -1.88]}