Question

20. Find the equation of the circle passing through the points (5,-8) (2,-9),(2,1)​

Answer 1

$\large\underline{\bold{Solution-}}$

$\sf \: Let \: us \: consider \: the \: equation \: of \: circle \: as$

$\sf \: {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 - - - (1)$

Since, (1) passes through (5, - 8).

Therefore,

$\sf \: {(5)}^{2} + {( - 8)}^{2} + 2g \times (5) + 2f \times ( - 8) + c = 0$

$\sf \: 25 + 64 + 10g - 16f + c = 0$

$\bf\implies \:10g - 16f + c = - 89 - - - (2)$

Again, (1) passes through (2, - 9)

Therefore,

$\sf \: {(2)}^{2} + {( - 9)}^{2} + 2g \times (2) + 2f \times ( - 9) + c = 0$

$\sf \: 4 + 81 + 4g - 18f + c = 0$

$\bf\implies \:4g - 18f + c = - 85 - - - (3)$

Also, (1) passes through (2, 1)

Therefore,

$\sf \: {(2)}^{2} + {(1)}^{2} + 2g \times 2 + 2f \times 1 + c = 0$

$\sf \: 4 + 1 + 4g + 2f + c = 0$

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

Now,

Subtracting equation (4) from equation (3), we get

$\sf \: - 20f = - 80$

$\bf :\implies\:f \: = \: 4 - - - (5)$

Now,

Subtracting equation (3) from equation (2),we get

$\rm :\longmapsto\:6g + 2f = - 4$

$\rm :\longmapsto\:6g + 2 \times 4 = - 4 \: \: \: \: \: \: \: \: \{ \because \: f \: = \: 4 \}$

$\rm :\longmapsto\:6g + 8 = - 4$

$\rm :\longmapsto\:6g = - 4 - 8$

$\rm :\longmapsto\:6g = - 12$

$\bf\implies \:g \: = \: - \: 2 - - - (6)$

On substituting the values of g and f in equation (5), we get

$\rm :\longmapsto\:4 \times ( - 2)+ 2 \times ( 4) + c = - 5$

$\rm :\longmapsto\:- 8 + 8+ c = - 5$

$\bf\implies \:c \: = \: - 5- - - (7)$

Now, Substitute the values of g, f and c in equation (1), the required equation of circle is

$\bf :\longmapsto\: {x}^{2} + {y}^{2} - 4x + 8y - 5 = 0$

Additional Information :-

$\sf \: Let \: us \: consider \: the \: equation \: of \: circle \: as$

$\sf \: {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 \: then$

$\rm :\longmapsto\:Centre = ( - g, , - f)$

and

$\rm :\longmapsto\:radius \: = \: \sqrt{ {g}^{2} + {f}^{2} - c }$