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?

3 Answers
May 15, 2018

(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

May 19, 2018

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.

May 27, 2018

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!