InterviewSolution
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InterviewSolution
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Tangents drawn from the point (4, 3) to the circle `x^(2)+y^(2)-2x-4y=0` are inclined at an angleA. `pi//6`B. `pi//4`C. `pi//3`D. `pi//2` |
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Answer» Correct Answer - D Using `SS_(1)=T^(2)`, the combined equation of the tangents drawn from (4, 3) to the circle `x^(2)+y^(2)-2x-4y=0`, is `(x^(2)+y^(2)-2x-4y)(16+9-8-12)` `{4x+2y-(x+y)-2(y+3)}^(2)` `rArr 5 (x^(2)+y^(2)-2x-4y)=(3x+y-10)^(2)` `rArr 4x^(2)-4y^(2)-50x+6xy+100=0 " " ...(i)` In this equation , we have Coeff. of `x^(2)+` Coeff. of `y^(2)=0`. Therefore, lines given by (i) are at right angle to each other. `ul("ALITER")` The equation of the given circle and its director circle are `(x-1)^(2)+(y-2)^(2)=5` and `(x-1)^(2)+(y-2)^(2)=10 " " (i)` Clearly, (4, 3) lies on (i). So, required angle is `pi//2`. |
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