Question

A circle has a center that falls on the line `y = 7/8x +1` and passes through `( 5 ,2 )` and `(3 ,6 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-8/3)^2+(y-10/3)^2=65/9`

Explanation:

Let `C` be the mid point of `A=(5,2)` and `B=(3,6)`

`C=((5+3)/2,(6+2)/2)=(4,4)`

The slope of `AB` is `=(6-2)/(3-5)=(4)/(-2)=-2`

The slope of the line perpendicular to `AB` is `=1/2`

The equation of the line passing trrough `C` and perpendicular to `AB` is

`y-4=1/2(x-4)`

`y=1/2x-2+4=1/2x+2`

The intersection of this line with the line `y=7/8x+1` gives the center of the circle.

`1/2x+2=7/8x+1`

`7/8x-1/2x=2-1`

`3/8x=1`

`x=8/3`

`y=1/2*(8/3)+2=10/3`

The center of the circle is `(8/3,10/3)`

The radius of the circle is

`r^2=(5-8/3)^2+(2-10/3)^2`

`=(7/3)^2+(-4/3)^2`

`=65/9`

The equation of the circle is

`(x-8/3)^2+(y-10/3)^2=65/9`

graph{((x-8/3)^2+(y-10/3)^2-65/9)(y-7/8x-1)(y-1/2x-2)=0 [-9, 11, -1.96, 8.04]}

Answer 2

`3x^2+3y^2-16x-20y+33=0`

Explanation:

Let the equation of circle be `x^2+y^2+2gx+2fy+c=0`, whose center is `(-g,-f)` and radius is `sqrt(g^2+f^2-c)`.

As the centre is on `y=7/8x+1`, we have `-f=-7/8g+1`

or `7g-8f=8` ......................(1)

It also passes through `(5,2)` and `(3,6)`, we have

`5^2+2^2+10g+4f+c=0` or `10g+4f+c+29=0` ......................(2)

`3^2+6^2+6g+12f+c=0` or `6g+12f+c+45=0` ......................(3)

Subtracting (2) from (3), we get

`-4g+8f+16=0` ......................(4)

Adding (1) and (4), we have `3g=-8` or `g=-8/3`

and puuting this in (1), we get `-8f=8-7(-8/3)=8+56/3=80/3`

Hence `f=-10/3` and then putting `g` and `f` in (2), we get

`10(-8/3)+4(-10/3)+c+29=0` or `-80/3-40/3+c+29=0`

i.e. `c=-29+120/3=11`

Hence, equaton of circle is

`x^2+y^2-16/3x-20/3y+11=0`

or `3x^2+3y^2-16x-20y+33=0`

graph{(3x^2+3y^2-16x-20y+33)((x-5)^2+(y-2)^2-0.01)((x-3)^2+(y-6)^2-0.01)(y-(7x)/8-1)=0 [-7.83, 12.17, -1.94, 8.06]}