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Equation of the tangent to the circle at the point (1, -1) whose centre is the point of intersection of the straight lines x-y=1 and 2x+y-3=0, isA. 3x-y-4=0B. x+4y+3=0C. x-3y-4=0D. 4x+y-3=0 |
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Answer» Correct Answer - B Given lines x-y-1=0 and 2x+y-3=0 intersect at C(4/3, 1/3). The equation of the circle passing through P(1, -1) and having centre at C(4/3, 1/3), is `(x-(4)/(3))^(2)+(y-(1)/(3))^(2)=(1-(4)/(3))^(2)+(-1-(1)/(3))^(2)` or, `x^(2)+y^(2)-(8)/(3)x-(2)/(3)y=0` The equation of the tangent to this circle at (1, -1), is `x-y-(4)/(3)(x+1)-(1)/(3)(y+1)=0` or, `x+4y+3=0` |
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