The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154sq. units.
The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154sq. units. Then the equation of the circle is
(a) x2 + y2 + 2x – 2y = 47
(b) x2 + y2 – 2x + 2y = 47
(c) x 2 + y2 – 2x + 2y = 62
(d) x2 + y2 + 2x – 2y = 62.
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(b) : Coordinate of centre may be (1, –1) or (–1, 1) but 1, –1 satisfies the given equations of diameter, so choices (a) and (d) are out of court. Again πR2 = 154, R2 = 49 ∴ R = 7
∴ Required equation of circle be
(x – 1) 2 + (y + 1) 2 = 49
=> x2 + y2 – 2x + 2y = 47
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