Question

Find the equation of a circle passing point (5,-8) ,(2,-9) and (2,1)

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

The general Equation of a circle is given by

$\implies \boxed{ x {}^{2} + y {}^{2} + 2gx + 2fy + c = 0}$

Now the points (5,-8) ,(2,-9) and (2,1)are on the circle .....

Satisfying point (5,-8)

$\implies (5) {}^{2} + ( - 8) {}^{2} + 2g(5) + 2f( - 8) + c = 0 \\ \implies 10g - 16f + c = - 89 \:</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>i)</strong><strong>$

Satisfying point (2,-9)

$= > 2 {}^{2} + ( - 9) {}^{2} + 2g(2) + 2f( - 9) + c = 0 \\ = > 4g - 18f + c = - 13</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>ii</strong><strong>)</strong><strong>$

Satisfying point (2,1)

$= > 2 {}^{2} + 1 {}^{2} + 2g(2) + 2f(1) + c = 0 \\ = > 4g + 2f + c = - 5</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>(</strong><strong>iii</strong><strong>)</strong><strong>$

Solving of g,f and c

from (i)&(ii)

10g - 16f + c = - 89

4g - 18f + c = - 13

____________________

@6g+2f= -76

=>3g+f= -38........(iv)

from (i)

10g - 16f + c = - 89

=>c=16f-10g-89

putting this in (iii)...

4g+2f+16f-10g-89=-5

=>-8g+18f=84

=>-4g+9f=42.........(v)

from equation (iv)&(v)

doing [(i)×9-(ii)×1], we get

=>27g+9f+4g-9f=-38-42

=>36g=-80

=>g=-80/36=-20/9

therefore...

f=42+4(-20/9)=42-(80/9)=298/9

and

c=-5-4(-20/9)-2(298/9)=-157/3

putting the values of g,f and c in the general form

=>x²+y²+2(-20/9)x+2(298/9)y+(-157/3)=0

$\implies\boxed{\red{9x{}^{2}+9y{}^{2}-40x+596y-471=0}}$

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