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

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 $x^{2} + y^{2} - 2 x + 2 y = 47$
Centre of circle = Point of intersection of diameters = (1, –1)
Now area = $154 \Rightarrow π r^{2} = 154 \Rightarrow r = 7$

Hence the equation of required circle is $(x - 1)^{2} + (y + 1)^{2} = 7^{2} \Rightarrow x^{2} + y^{2} - 2 x + 2 y = 47.$

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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

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

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

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