InterviewSolution
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InterviewSolution
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The locus of the mid points of the chords of the circle `x^2+y^2+4x-6y-12=0` which subtend an angle of `pi/3`radians at its circumference is:(A) `(x-2)^2+(y+3)^2=6.25` (B) `(x+2)^2+(y-3)^2=6.25`(C) `(x+2)^2+(y-3)^2=18.75` (D) `(x+2)^2+(y+3)^2=18.75` |
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Answer» We can draw the figure from the given details. Please refer to the video for diagram. Equation of the given circle is is, `x^2+y^2+4x-6y-12 = 0` Comparing this equation with the standard circle equation, `x^2+y^2+2gx+2fy+c = 0`, we get `g = 2,f = -3, c = -12` Now, centre of circle will be `(-g,-f) = (-2,3)` And, radius AC will be `sqrt(g^2+f^2-c) = sqrt(4+9+12) = 5` Now, we are given ,`/_AQB = 60^@` `/_ACB` will be twice of it, as C is centre of the circle. `/_ACB = 2**60^@ = 120^@` Now, if `P(alpha,beta)` is the midpoint of the chord, then `/_ACP = 60^@` So, CP will be `/_ACcos60^@ = 5/2` Now, we can say that `CP^2 = ((alpha - (-2))^2 + (beta -3)^2)` `=>(alpha+2)^2+(beta-3)^2 = (2.5)^2` `=>(alpha+2)^2+(beta-3)^2 = 6.25` If we replace (alpha,beta) with (x,y), then, `(x+2)^2+(y-3)^2 = 6.25` that is the required equation. |
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