The lines $2 x - 3 y = 5$ & $3 x - 4 y = 7$ are diameters of a circle of area $154$ sq units Then the equation of the circle is

x^{2} + y^{2} + 2 x - 2 y = 62

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x^{2} + y^{2} - 2 x + 2 y = 47

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x^{2} + y^{2} + 2 x - 2 y = 47

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

x^{2} + y^{2} - 2 x + 2 y = 62

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Solution

The correct option is B $x^{2} + y^{2} - 2 x + 2 y = 47$
Intersection of any 2 diameters gives us the center

$2 x - 3 y = 5 - e q .1$

$3 x - 4 y = 7 - e q .2$

$3 \times e q .1 - 2 \times e q .2$

$⟹ - 9 y + 8 y = 15 - 14$

$⟹ y = - 1 ⟹ x = 1$

Center = (1,-1)

And given A = 154

$⟹ π r^{2} = 154$

$r^{2} = \frac{154}{22} \times 7 = 49$

$r = 7$

Equation of circle is

$(x - 1)^{2} + (y + 1)^{2} = 7^{2}$

$⟹ x^{2} + y^{2} - 2 x + 2 y - 47 = 0$

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Similar questions

The lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 square units. The equation of the circle is

P (2, 2)

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

A

B .

△ P A B

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

90 °

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