Question

find the equation of the circle which passes through (4,1),(6,5) and having centre on 4x+3y-24=0

Answer 1

$Let \: the \:equation \:of \:the \:circle \:be \\\pink { (x-h)^{2} + (y-k)^{2} = r^{2}} \: ---(1)$

$i) Since \: circle \: passes \: through \:(4,1) \\it \: will \:satisfy \: equation \: of \:circle$

$put \: x = 4 \: and \: y = 1 \: in \: (1)$

$\implies (4-h)^{2} + ( 1-k)^{2} = r^{2}$

$\implies 4^{2} + h^{2}-2\times 4 \times h + 1^{2} + k^{2} - 2 \times 1 \times k = r^{2}$

$\implies 16+h^{2} - 8h + 1+k^{2} - 2k = r^{2}$

$\implies h^{2}+k^{2}-8h-2k+17=r^{2} \: --(2)$

$ii) Since \: circle \: passes \: through \:(6,5) \\it \: will \:satisfy \: equation \: of \:circle$

$put \: x = 6\: and \: y = 5\: in \: (1)$

$\implies (6-h)^{2} + ( 5-k)^{2} = r^{2}$

$\implies 6^{2} + h^{2}-2\times 6\times h + 5^{2} + k^{2} - 2 \times 5\times k = r^{2}$

$\implies 36+h^{2} - 12h + 25+k^{2} - 10k = r^{2}$

$\implies h^{2}+k^{2}-12h-10k+61=r^{2} \: --(3)$

$iii ) Since , centre \: of \: circle \:(h,k) \:lie \\on \:the \: line \: 4x + 3y - 24 = 0$

$\implies 4h + 3k = 24 \: ---(4)$

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

$h^{2}+k^{2}-8h-2k+17-(h^{2}+k^{2}-12h-10k+61) = r^{2} - r^{2}$

$\implies h^{2}+k^{2}-8h-2k+17- h^{2}-k^{2}+12h+10k-61= 0$

$\implies 4h + 8k - 44 = 0$

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

$\implies h + 2k - 11 = 0$

$\implies h = -2k + 11 \: ---(5)$

/* Substitute (5) in equation (4) , we get */

$\implies 4(-2k+11) + 3k = 24$

$\implies -8k + 44 + 3k = 24$

$\implies -5k = 24 - 44$

$\implies -5k = - 20$

$\implies k = \frac{- 20}{-5}$

$\implies k = 4$

/* put k = 4 in equation (5) , we get */

$h = -2\times 4 + 11$

$\implies h = -8 + 11$

$\implies h = 3$

$Put \:h = 3 \: and \: k = 4 \: in \: equation \:(2) ,\\we \:get$

$3^{2}+4^{2}-8\times 3 - 2\times 4 + 17 = r^{2}$

$\implies 9 + 16 - 24 - 8 + 17 = r^{2}$

$\implies 10 = r^{2}$

$Put \:h = 3 \: and \: k = 4 \: and \: r^{2} = 10 \:in\\ equation \:(1) ,\:we \:get$

$\implies ( x - 3 )^{2} + ( y - 4 )^{2} = 10$

$\implies x^{2} + 3^{2} - 2\times x \times 3 + y^{2} + 4^{2} - 2\times y \times 4 = 10$

$\implies x^{2} + 9 - 6x + y^{2} + 16 - 8y = 10$

$\implies x^{2} + y^{2} -6x -8y + 25-10 = 0$

$\implies x^{2} + y^{2} -6x -8y + 15 = 0$

Therefore.,

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

$\green { x^{2} + y^{2} -6x -8y + 15 = 0 }$

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Answer 2

Answer:

Step-by-step explanation:

Let required equation be x²+y²+2gx+2fy+c=0

Substitute A(4,1) in equation 1

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