If the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

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Solution

Area of given circle is it 154
$π r^{2} = 154 \frac{22}{7} r^{2} = 154 r^{2} = 154 \times \frac{7}{22} r^{2} - 49 r = 7$
The intersection point of 2x - 3y it 5 and 3x -4y = 7 is
The centre of the circle. Solving simultaneous equations
$2 x - 3 y = 5 a n d 3 x - 4 y = 7$
we get, Centre of circle as (1, -1)
Equation of circle with centre (1, -1) and radius = 7 is
$(x - 1)^{2} + (y + 1)^{2} = 7^{2} x^{2} - 2 x + 1 + y^{2} + 2 y + 1 = 49 x^{2} - 2 x + y^{2} + 2 y = 47$

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Similar questions

2 x - 3 y - 5 = 0

3 x - 4 y = 7

154

Find the equation of circle, if the lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle of area 154 sq units.

2 x - 3 y - 5 = 0

3 x - 4 y = 7

154 s q u n i t s

2 x - 3 y = 5

3 x - 4 y = 7

154

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