The equation of a circle of area 154 sq.units, two of whose diameters are $2 x - 3 y + 12 = 0$ and $x + 4 y - 5 = 0$ , is :

x^{2} + y^{2} - 6 x + 4 y - 36 = 0

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x^{2} + y^{2} - 6 x - 4 y - 36 = 0

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

x^{2} + y^{2} + 6 x - 4 y - 36 = 0

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x^{2} + y^{2} + 6 x + 4 y + 36 = 0

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Solution

The correct option is C $x^{2} + y^{2} + 6 x - 4 y - 36 = 0$
Given,

$2 x - 3 y + 12 = 0$ ...(1)

$x + 4 y - 5 = 0$ ...(2)

$(1) - 2 (2)$ gives,

$y = 2$

$x = - 3$

Center of the circle $(- 3, 2)$

Area of circle $= 154$

$π r^{2} = 154$

$r^{2} = 154 \times \frac{7}{22}$

$r = 7$

Hence the equation of circle is,

$(x - (- 3))^{2} + (y - 2)^{2} = 7^{2}$

$(x + 3)^{2} + (y - 2)^{2} = 7^{2}$

solving the above equation, we get,

$x^{2} + 6 x + y^{2} + 13 - 4 y = 49$

$x^{2} + 6 x + y^{2} - 4 y - 36 = 0$

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Similar questions

Find the equation of director circle of the circle whose diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area is 154 square units.

S_{1} \equiv x^{2} + y^{2} - 4 y - 5 = 0

S_{2} \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0

(1, - 2)

x^{2} + y^{2} = 1

(3, 2)

Equation of the circle on the latusrectum of $y^{2} = 8 x$ as ends of diameter is

x^{2} + y^{2} + 8 x - 4 y + c = 0

x^{2} + y^{2} + 2 x + 4 y - 11 = 0

x^{2} + y^{2} - 6 x + 8 y + k = 0

k =

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