Let {{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

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Answer: B. 8 {{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}} = bx\,dy + cy\,dy + a\,dy = ax\,dx - by\,dx + a\,dx = cy\,dy + a\,dy - ax\,dx - a\,dx + b(x\,dy + y\,dx) = 0 = c\int {y\,dy + a\int {x\,dx - a\int {dx + b\int {d(xy) = 0} } } } = {{c{y^2}} \over 2} + ay - {{a{x^2}} \over 2} - ax + bxy = k = a{x^2} - c{y^2} + 2ax - 2ay - 2bxy = k Above equation is circle \Rightarrow a = - c and b = 0 a{x^2} + a{y^2} + 2ax - 2ay = k \Rightarrow {x^2} + {y^2} + 2x - 2y = \lambda \,\,\,\,\,\,\,\left[ {\lambda = {k \over a}} \right] Passes through (2, 5) 4 + 25 + 4 - 10 = \lambda \Rightarrow \lambda = 23 Circle \equiv {x^2} + {y^2} + 2x - 2y - 23 = 0 Centre (-1, 1) r = \sqrt {{{( - 1)}^2} + {1^2} + 23} = 5 Shortest distance of (11,6) = \sqrt {{{12}^2} + {5^2}} - 5 = 13 - 5 = 8

Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :

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