Question

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

Answer 1

`(x + 8)^2 + (y-1)^2 = 170`

Explanation:

I don't trust the "12 minutes ago" without a comment. Where'd they go?

We call the center `(x,y)` and equate the squared radii, brashly writing the equation:

`(x - 5)^2 + ( (5/8 x + 6) - 2)^2 = (x - 3)^2 + ( (5/8 x + 6) - 8)^2`

`(x - 5)^2 + (5/8 x + 4)^2 = (x - 3)^2 + ( 5/8 x - 2)^2`

Let's multiply by `8^2` to clear the fractions:

`64 (x - 5)^2 + (5 x + 32)^2 = 64(x - 3)^2 + ( 5 x - 16)^2`

Let's expand. We see the squared terms are the same on each side so I won't bother to write them. Once that's done it looks like there's a factor of 32 that can be cancelled too:

`32( -20 x + 50 + 10 x + 32) = 32( - 12 x +18 - 5 x + 8)`

`7 x = -56`

`x = -8`

`y = 5/8 x + 6 = 1`

Squared radius `(-8 -5)^2 + (1 - 2)^2 = 170`

Equation:

`(x + 8)^2 + (y-1)^2 = 170`

Check:

`(3+8)^2 + (8-1)^2 = 11^2+7^2=121+49=170 quad sqrt`

Answer 2

`S : x^2+y^2+16x-2y-105=0`.

Explanation:

Suppose that the reqd. eqn. of the circle `S` is given by,

`S : x^2+y^2+2gx+2fy+c=0`.

The centre `C` of `S` is `C(-g,-f) in" the line : "y=5/8*x+6`.

`:. -f=-5/8*g+6, or, f=5/8g-6.........................(1)`.

`(5,2) in S. :. 5^2+2^2+10g+4f+c=0, or,`

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

Likewise, `(3,8) in S. :. 73+6g+16f+c=0.................(3)`.

`(3)-(2) rArr 44-4g+12f=0 rArr 11-g+3f=0.........(4)`.

`(1) & (4) rArr 11-g+3(5/8g-6)=0`.

`:. -7+7/8g=0 rArr g=8........................................(star1)`.

`:. f=5/8g-6=5/8*8-6=-1............................(star2)`.

With `g=8,f=-1 and (2)`, we get,

`29+80-4+c=0 rArr c=-105.............................(star3)`.

`(star1), (star2) and (star3)` give us,

`S : x^2+y^2+16x-2y-105=0`.

Answer 3

`x^2+y^2+16x-2y-105=0`,

Explanation:

Here is a Third Method to find the eqn. of the circle in Question.

The eqn. of a circle having `(5,2) and (3,8)` as extremities of its

diameter is given by, `s : (x-5)(x-3)+(y-2)(y-8)=0`.

`:. s : x^2+y^2-8x-10y+31=0`.

The eqn. of the chord joining `(5,2) and (3,8)` is given by,

`l : |(x,y,1),(5,2,1),(3,8,1)|=0`,

`i.e., -6x-2y+34=0 or, l : 3x+y-17=0`.

Now, since the reqd. circle `S` passes through the points of

intersection of `s and l`, we may take, `S : s+lambdal=0, or,`

`S : x^2+y^2-8x-10y+31+lambda(3x+y-17)=0`,

`i.e., S : x^2+y^2+(3lambda-8)x+(lambda-10)y+31-17lambda=0`.

Clearly, the centre `C` of `S` is,

`C(4-3/2lambda,5-lambda/2)`, which lies on `y=5/8x+6`.

`:. 5-lambda/2=5/8(4-3/2lambda)+6`.

`:. lambda=8`.

Hence, the eqn. of reqd. circle is

`S : x^2+y^2-8x-10y+31+8(3x+y-17)=0`,

`i.e., S : x^2+y^2+16x-2y-105=0`, as before!