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

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

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Answer 1 · 2017-07-02T08:20:03

The equation of the circle is ${(x - \frac{8}{3})}^{2} + {(y - \frac{10}{3})}^{2} = \frac{65}{9}$ $(x-8/3)^2+(y-10/3)^2=65/9$

Explanation:

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

$C = (\frac{5 + 3}{2}, \frac{6 + 2}{2}) = (4, 4)$ $C=((5+3)/2,(6+2)/2)=(4,4)$

The slope of $A B$ $AB$ is $= \frac{6 - 2}{3 - 5} = \frac{4}{- 2} = - 2$ $=(6-2)/(3-5)=(4)/(-2)=-2$

The slope of the line perpendicular to $A B$ $AB$ is $= \frac{1}{2}$ $=1/2$

The equation of the line passing trrough $C$ $C$ and perpendicular to $A B$ $AB$ is

$y - 4 = \frac{1}{2} (x - 4)$ $y-4=1/2(x-4)$

$y = \frac{1}{2} x - 2 + 4 = \frac{1}{2} x + 2$ $y=1/2x-2+4=1/2x+2$

The intersection of this line with the line $y = \frac{7}{8} x + 1$ $y=7/8x+1$ gives the center of the circle.

$\frac{1}{2} x + 2 = \frac{7}{8} x + 1$ $1/2x+2=7/8x+1$

$\frac{7}{8} x - \frac{1}{2} x = 2 - 1$ $7/8x-1/2x=2-1$

$\frac{3}{8} x = 1$ $3/8x=1$

$x = \frac{8}{3}$ $x=8/3$

$y = \frac{1}{2} \cdot (\frac{8}{3}) + 2 = \frac{10}{3}$ $y=1/2*(8/3)+2=10/3$

The center of the circle is $(\frac{8}{3}, \frac{10}{3})$ $(8/3,10/3)$

The radius of the circle is

$r^{2} = {(5 - \frac{8}{3})}^{2} + {(2 - \frac{10}{3})}^{2}$ $r^2=(5-8/3)^2+(2-10/3)^2$

$= {(\frac{7}{3})}^{2} + {(- \frac{4}{3})}^{2}$ $=(7/3)^2+(-4/3)^2$

$= \frac{65}{9}$ $=65/9$

The equation of the circle is

${(x - \frac{8}{3})}^{2} + {(y - \frac{10}{3})}^{2} = \frac{65}{9}$ $(x-8/3)^2+(y-10/3)^2=65/9$

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 · 2017-07-02T08:33:43

$3 x^{2} + 3 y^{2} - 16 x - 20 y + 33 = 0$ $3x^2+3y^2-16x-20y+33=0$

Explanation:

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

As the centre is on $y = \frac{7}{8} x + 1$ $y=7/8x+1$ , we have $- f = - \frac{7}{8} g + 1$ $-f=-7/8g+1$

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

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

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

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

Subtracting (2) from (3), we get

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

Adding (1) and (4), we have $3 g = - 8$ $3g=-8$ or $g = - \frac{8}{3}$ $g=-8/3$

and puuting this in (1), we get $- 8 f = 8 - 7 (- \frac{8}{3}) = 8 + \frac{56}{3} = \frac{80}{3}$ $-8f=8-7(-8/3)=8+56/3=80/3$

Hence $f = - \frac{10}{3}$ $f=-10/3$ and then putting $g$ $g$ and $f$ $f$ in (2), we get

$10 (- \frac{8}{3}) + 4 (- \frac{10}{3}) + c + 29 = 0$ $10(-8/3)+4(-10/3)+c+29=0$ or $- \frac{80}{3} - \frac{40}{3} + c + 29 = 0$ $-80/3-40/3+c+29=0$

i.e. $c = - 29 + \frac{120}{3} = 11$ $c=-29+120/3=11$

Hence, equaton of circle is

$x^{2} + y^{2} - \frac{16}{3} x - \frac{20}{3} y + 11 = 0$ $x^2+y^2-16/3x-20/3y+11=0$

or $3 x^{2} + 3 y^{2} - 16 x - 20 y + 33 = 0$ $3x^2+3y^2-16x-20y+33=0$

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]}

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

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