The equation of the circle having centre $(1, - 2)$ and passing through the point of intersection of the lines $3 x + y = 14$ and $2 x + 5 y = 18$ is

x^{2} + y^{2} - 2 x + 4 y - 20 = 0

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x^{2} + y^{2} - 2 x - 4 y - 20 = 0

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x^{2} + y^{2} + 2 x - 4 y - 20 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + 2 x + 4 y - 20 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$
The circle passes through intersection of $3 x + y = 14$ and $2 x + 5 y = 18$ .
$∴ 5 (3 x + y - 14) - 2 x - 5 y + 18 = 0 ⟹ 15 x - 70 - 2 x + 18 = 0 ⟹ x = 4 ⟹ y = 2$
Hence, equation of circle is
$(x - 1)^{2} + (y + 2)^{2} = (4 - 1)^{2} + (2 + 2)^{2}$
$∴ x^{2} - 2 x + 1 + y^{2} + 4 y + 4 = 25$
$∴ x^{2} + y^{2} - 2 x + 4 y - 20 = 0$
This is the required equation.

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Similar questions

Q. The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 3x + 5y = 18 is
(a) x² + y² – 2x + 4y – 20 = 0
(b) x² + y² – 2x – 4y – 20 = 0
(c) x² + y² + 2x – 4y – 20 = 0
(d) x² + y² + 2x + 4y – 20 = 0

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(a) x² + y² − 2x − 4y + 4 = 0
(b) x² + y² + 2x + 4y − 4 = 0
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$x^{2} - y^{2} - 2 x + 4 y - 3 = 0,$ is

The radius of the circle S is same as the radius of $x^{2} + y^{2} - 2 x + 4 y - 11 = 0$ and the centre of S is the centre of $x^{2} + y^{2} - 2 x - 4 y + 11 = 0$ . Find the equation of S.

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The equation of the circle having centre (1, −2) and passing through the point of intersection of the lines 3x+y=14 and 2x+5y=18 is

The equation of the circle having centre $(1, - 2)$ and passing through the point of intersection of the lines $3 x + y = 14$ and $2 x + 5 y = 18$ is