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Find the locus of center of circle of radius 2 units, if intercept cuton the x-axis is twice of intercept cut on the y-axis by the circle. |
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Answer» Correct Answer - `4x^(2)-y^(2)=12` Let the circle be `x^(2)+y^(2)+2gx+2fy+c=0` Therefore , the intercept on the x-axis is `2 sqrt(g^(2)-c)` Also, the intercept on the y-axis is `2 sqrt(f^(2)-c)` According to the question, `sqrt(g^(2)-c)=2sqrt(f^(2)-c)` or `g^(2)-4f^(2)= -3c` (1) Given, radius `=2` `:. g^(2)+f^(2)-c=4` (2) Eliminating c from (1) and (2), we get `4g^(2)-f^(2)=12` Therfore, the locus is `4x^(2)-y^(2)=12` |
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