A circle has a center that falls on the line $y = \frac{7}{4} x + 4$ $y = 7/4x +4$ and passes through $(4, 7)$ $( 4 ,7 )$ and $(7, 5)$ $(7 ,5 )$ . What is the equation of the circle?

Question

A circle has a center that falls on the line $y = \frac{7}{4} x + 4$ $y = 7/4x +4$ and passes through $(4, 7)$ $( 4 ,7 )$ and $(7, 5)$ $(7 ,5 )$ . What is the equation of the circle?

1 Answer

Impact of this question

1802 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-28T05:41:09

$2 x^{2} + 2 y^{2} + 100 x + 159 y - 1643 = 0$ $2x^2 + 2y^2 + 100x + 159y - 1643 = 0$

Explanation:

General equation of circle can be represented as $x^{2} + y^{2} - 2 h x - 2 k y + c = 0$ $x^2+y^2-2hx-2ky+c=0$ -> $E q u a t i o n 1$ $Equation 1$

where $(h, k)$ $(h,k)$ represents the center of the circle , $r$ $r$ - radius of the circle, $c = h^{2} + k^{2} - r^{2}$ $c = h^2 + k^2 - r^2$

As we know the given points $(4, 7) and (7, 5)$ $(4,7) and (7,5)$ lie on the circle, they must satisfy Equation 1.

Putting the given points $(4, 7) and (7, 5)$ $(4,7) and (7,5)$ in the above equation of circle (1) , we get :

$4^{2} + 7^{2} - 2 h \cdot 4 - 2 k \cdot 7 + c = 0$ $4^2+7^2-2h*4-2k*7+c=0$

=> $- 8 h - 14 k + c = - 65$ $-8h - 14k + c = -65$ =>

$8 h + 14 k - c = 65$ $8h + 14k - c = 65$ -> $E q u a t i o n 2$ $Equation 2$

$7^{2} + 5^{2} - 2 h \cdot 7 - 2 k \cdot 5 + c = 0$ $7^2+5^2-2h*7-2k*5+c=0$

=> $- 14 h - 10 k + c = - 74$ $-14h - 10k +c=-74$ =>

$14 h + 10 k - c = 74$ $14h + 10k - c=74$ -> $E q u a t i o n 3$ $Equation 3$

Since the centre of the circle is $(h, k)$ $(h,k)$ lies on the line $y = \frac{7}{4} x + 4$ $y = 7/4x + 4$ , $(h, k)$ $(h,k)$ must satisfy the equation of line.

$k = \frac{7}{4} \cdot (h) + 4$ $k = 7/4*(h) + 4$ =>

$7 h - 4 k = - 16$ $7h - 4k = -16$ -> $E q u a t i o n 4$ $Equation 4$

Subtract Equation 2 and 3

=> $(14 h - 8 h) + (10 k - 14 k) + (- c + c) =$ $(14h - 8h) + (10k - 14k) + (-c+c) =$ 74 - 65#

$6 h - 4 k = 9$ $6h -4k = 9$ -> $E q u a t i o n 5$ $Equation 5$

Add Equation 4 and 5

=> $(- 7 h + 6 h) + (4 k - 4 k) = (16 + 9)$ $(-7h+6h) + (4k -4k) = (16+9)$ -> $E q u a t i o n 5$ $Equation 5$

$- h = 25 \Rightarrow h = - 25$ $-h = 25 => h= -25$

$- 4 k = 9 - (6 \cdot (- 25)$ $-4k= 9 - (6*(-25)$

$k = - \frac{159}{4}$ $k = -159/4$

Solving for c in Equation 3, we get $c = 14 \cdot (- 25) + 10 \cdot (- \frac{159}{4}) - 74$ $c = 14*(-25) + 10*(-159/4) - 74$

$c = - \frac{3286}{4}$ $c= -3286/4$ = $- \frac{1643}{2}$ $-1643/2$

Circle equation is:

$x^{2} + y^{2} - 2 (- 25) x - 2 (- \frac{159}{4}) y - \frac{1643}{2} = 0$ $x^2+y^2-2(-25)x-2(-159/4)y- 1643/2=0$

$x^{2} + y^{2} + 50 x - (- \frac{159}{2}) y - \frac{1643}{2} = 0$ $x^2+y^2+50x-(-159/2)y - 1643/2=0$

$2 x^{2} + 2 y^{2} + 100 x + 159 y - 1643 = 0$ $2x^2 + 2y^2 + 100x + 159y - 1643 = 0$

Science

Math

Humanities

... and beyond

A circle has a center that falls on the line $y = \frac{7}{4} x + 4$ $y = 7/4x +4$ and passes through $(4, 7)$ $( 4 ,7 )$ and $(7, 5)$ $(7 ,5 )$ . What is the equation of the circle?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A circle has a center that falls on the line y=74x+4y = 7/4x +4 and passes through (4,7) ( 4 ,7 ) and (7,5)(7 ,5 ). What is the equation of the circle?

1 Answer

Explanation:

Related questions

Impact of this question

A circle has a center that falls on the line $y = \frac{7}{4} x + 4$ $y = 7/4x +4$ and passes through $(4, 7)$ $( 4 ,7 )$ and $(7, 5)$ $(7 ,5 )$ . What is the equation of the circle?