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The lines `2x-3y-5=0` and `3x-4y=7` are diameters of a circle of area `154(=49 pi)` sq. units, then the equation of the circle isA. `x^(2)+y^(2)+2x-2y-62=0`B. `x^(2)+y^(2)+2x-2y-47=0`C. `x^(2)+y^(2)-2x+2y-47=0`D. `x^(2)+y^(2)-2x+2y-62=0` |
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Answer» Correct Answer - C The centre of the required circle lies at the intersection of `2x-3y-5=0` and `3x-4y-7=0`. Thus, the coordinates of the centre are (1, -1). Let r be the radius of the circle. Then, by hypothesis, we have `pi r^(2)=154 rArr (22)/(7) r^(2)=154 rArr r=7` Hence, the equation of the required circle is `(x-1)^(2)+(y+1)^(2)=7^(2) rArr x^(2)+y^(2)-2x+2y-47=0` |
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