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

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

1 Answer

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Answer 1 · 2016-10-19T13:13:43

Equation of circle is $16 x^{2} + 16 y^{2} + 27 x - 206 y + 123 = 0$ $16x^2+16y^2+27x-206y+123=0$

Explanation:

As the circle passes through points $(5, 7)$ $(5,7)$ and $(3, 2)$ $(3,2)$ , its center is equidistant from these two points and hence lies on perpendicular bisector of points $(5, 7)$ $(5,7)$ and $(3, 2)$ $(3,2)$ , whose equation is given by

${(x - 5)}^{2} + {(y - 7)}^{2} = {(x - 3)}^{2} + {(y - 2)}^{2}$ $(x-5)^2+(y-7)^2=(x-3)^2+(y-2)^2$

or $x^{2} - 10 x + 25 + y^{2} - 14 y + 49 = x^{2} - 6 x + 9 + y^{2} - 4 y + 4$ $x^2-10x+25+y^2-14y+49=x^2-6x+9+y^2-4y+4$

or $- 10 x + 25 - 14 y + 49 + 6 x - 9 + 4 y - 4 = 0$ $-10x+25-14y+49+6x-9+4y-4=0$

or $- 4 x - 10 y + 61 = 0$ $-4x-10y+61=0$ or $4 x + 10 y = 61$ $4x+10y=61$ ......................(A)

As center also lies on $y = \frac{2}{3} x + 7$ $y=2/3x+7$ , putting this in (A)

$4 x + 10 (\frac{2}{3} x + 7) = 61$ $4x+10(2/3x+7)=61$ or $12 x + 20 x + 210 = 183$ $12x+20x+210=183$

or $32 x = 183 - 210 = - 27$ $32x=183-210=-27$ and $x = - \frac{27}{32}$ $x=-27/32$ and

$y = \frac{2}{3} \times (- \frac{27}{32}) + 7 = - \frac{9}{16} + 7 = \frac{103}{16}$ $y=2/3xx(-27/32)+7=-9/16+7=103/16$

Hence, center is $(- \frac{27}{32}, \frac{103}{16})$ $(-27/32,103/16)$ and radius squared is ${(- \frac{27}{32} - 3)}^{2} + {(\frac{103}{16} - 2)}^{2}$ $(-27/32-3)^2+(103/16-2)^2$

And equation of circle is

${(x + \frac{27}{32})}^{2} + {(y - \frac{103}{16})}^{2} = {(- \frac{27}{32} - 3)}^{2} + {(\frac{103}{16} - 2)}^{2}$ $(x+27/32)^2+(y-103/16)^2=(-27/32-3)^2+(103/16-2)^2$

or $x^{2} + \frac{27}{16} x + {(\frac{27}{32})}^{2} + y^{2} - \frac{103}{8} y + {(\frac{103}{16})}^{2} = {(\frac{123}{32})}^{2} + {(\frac{71}{16})}^{2}$ $x^2+27/16x+(27/32)^2+y^2-103/8y+(103/16)^2=(123/32)^2+(71/16)^2$

Multiplying by $32^{2} = 1024$ $32^2=1024$ , we get

$1024 x^{2} + 27 \times 64 x + 729 + 1024 y^{2} - 103 \times 128 y + 206^{2} = 123^{2} + 142^{2}$ $1024x^2+27xx64x+729+1024y^2-103xx128y+206^2=123^2+142^2$

or $1024 x^{2} + 1728 x + 729 + 1024 y^{2} - 13184 y + 42436 = 15129 + 20164$ $1024x^2+1728x+729+1024y^2-13184y+42436=15129+20164$

or $1024 x^{2} + 1024 y^{2} + 1728 x - 13184 y + 7872 = 0$ $1024x^2+1024y^2+1728x-13184y+7872=0$ and dividing by $8$ $8$

$128 x^{2} + 128 y^{2} + 216 x - 1648 y + 984 = 0$ $128x^2+128y^2+216x-1648y+984=0$ and dividing again by $8$ $8$

$16 x^{2} + 16 y^{2} + 27 x - 206 y + 123 = 0$ $16x^2+16y^2+27x-206y+123=0$

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

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Explanation:

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A circle has a center that falls on the line y = 2/3x +7 y=23x+7y = 2/3x +7 and passes through (5 ,7 )(5,7)(5 ,7 ) and (3 ,2 )(3,2)(3 ,2 ). What is the equation of the circle?

1 Answer

Explanation:

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