Question

Find the equation of circle passing through (2,1),(5,5), (-6,7)

Answer 1

Given: Three points  (2,1), (5,5), (-6,7).

To find: The equation of circle.

Solution:

• Now the general equation of the circle is:

             x² + y² + 2gx + 2fy + c = 0 ……………. (i)

• So according to the problem, the above equation of the circle passes through the points (2,1), (5,5), (-6,7). Therefore, the equations will be:

            for (2,1): 4 + 1 + 4g + 2f + c = 0

             5 + 4g + 2f + c = 0  .....................(ii)

            for (5,5): 25 + 25 + 10g + 10f + c = 0

             50 + 10g + 10f + c = 0 .................(iii)

            for (-6,7): 36 + 49 - 12g + 14f + c = 0

             85 - 12g + 14f + c = 0  ...................(iv)

• Now iii - ii, we get:

             45 + 6g + 8f = 0 .................(v) Now iv - ii, we get:

             80 - 16g + 12f = 0 ...................(vi)

• Now 2(vi) - 3(v), we get:

             160 - 32g + 24f - 135 - 18g - 24f = 0

             25 -50g = 0

             g = 25/50

             g = 1/2

• Put g = 1/2 in vi, we get:

             80 - 16(1/2) + 12f = 0

             80 - 8 + 12f = 0

             12f = -72

             f = -72/12 = -6

• Now putting f and g in ii, we get:

             5 + 4(1/2) + 2(-6) + c = 0

             c = -5 - 2 + 12

             c = 12 - 7

             c = 5

• Therefore equation of circle is:

               x² + y² + x - 12y + 5 = 0

Answer:

         So the required equation of circle passing through (2,1), (5,5), (-6,7) is  x² + y² + x - 12y + 5 = 0

Answer 2

$Let \: the \: equation \:of \:the \: general \:form \\of \: the \: required \: circle \:be \\\green {x^{2}+y^{2}+2gx+2fy+c = 0 \: ---(1) }$

/* According to the problem , the above equation of the Circle passes through the points (2,1), (5,5) and (-6,7) */

$i ) Substitute \:(2,1) \: in \:equation \: (1)$

$\implies 2^{2}+1^{2}+2g\times 2+2f\times 1+c=0$

$\implies 4+1+4g+2f+c=0\\\implies 4g+2f+c+5= 0 \: ---(2)$

$ii ) Substitute \:(5,5) \: in \:equation \: (1)$

$\implies 5^{2}+5^{2}+2g\times 5+2f\times 5+c=0$

$\implies 25+25+10g+10f+c=0\\\implies 10g+10f+c+50 = 0 \: ---(3)$

$iii ) Substitute \:(-6,7) \: in \:equation \: (1)$

$\implies (-6)^{2}+7^{2}+2g\times (-6)+2f\times 7+c=0$

$\implies 36+49-12g+14f+c=0\\\implies -12g+14f+c+85 = 0 \: ---(4)$

/* Subtract equation (2) from equation (3) , we get */

$6g + 8f + 45 = 0 \: ---(5)$

/* Subtract equation (4) from equation (2) , we get */

$16g - 12f - 80 = 0$

/* Divide each term by 4, we get */

$\implies 4g - 3f - 20 = 0 \: --(6)$

/* Do 2 × (5) - 3× (6) , we get */

$25f + 150 = 0$

$\implies f = \frac{-150}{25}$

$\implies f = - 6\: ---(7)$

/* Put f = -6 in equation (5), we get */

$6g + 8(-6) + 45 = 0$

$\implies 6g - 48 + 45 = 0$

$\implies 6g - 3= 0$

$\implies g = \frac{3}{6} = \frac{1}{2} \:---(8)$

/* Put values of f and g in equation (2), we get */

$4\times \frac{1}{2} + 2\times (-6) + c + 5 = 0$

$\implies 2 - 12 + 5 + c = 0$

$\implies -5 + c = 0$

$\implies c = 5\: ---(9)$

/* Substituting the values of g, f and c in (2), we obtain the equation of the required circle as

$x^{2} + y^{2} +2\times \frac{1}{2} x + 2 \times (-6) y+ 5 = 0$

$\red{Required \:equation \: of \:the \:circle }$

$\green{ \: x^{2} + y^{2} + x - 12y + 5 = 0}$