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A circle passes through the points (3,4) and cuts the circle x^(2) + y^(2) = a^(2) orthogonally, the locus of its centre is a straight line. If the distance of this straight line from the origin is 25, then a^(2) is equal to |
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Answer» 250 Let `C (x_(1), y_(1))` be the CENTRE of the circle Circle are orthogonal to each other `implies d^(2) = r_(1)^(2) + r_(2)^(2)` `implies x_(1)^(2) + y_(1)^(2) = a^(2) + (x_(1) - 3)^(2) (y_(1) - 4)^(2)` `implies x_(1)^(2) + y_(1)^(2) = a^(2) + x_(1)^(2) - 6x_(1) + 9 + y_(1)^(2) - 8y_(1) + 16` `:.` Locus of `C (X_(1),y_(1))` is, `6x + 8y = a^(2) + 25` Distance from (0,0) to EQUATION (1) is `(a^(2) + 25)/(sqrt(64 + 36)) = 25` `implies a^(2) + 25 = 250` `:. a^(2) = 225` |
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