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The equation of the circle on the common chord of the circles `(x-a)^(2)+y^(2)=a^(2)` and `x^(2)+(y+b)^(2)=b^(2)` as diameter, isA. `x^(2)+y^(2)=2ab(bx+ay)`B. `x^(2)+y^(2)=bx+ay)`C. `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)`D. `(a^(2)+b^(2))(x^(2)+y^(2))=2(bx+ay)` |
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Answer» Correct Answer - C The equation of the common chord of the circles `(x-a)^(2)+y^(2)=a^(2) and x^(2)+(y+b)^(2)=b^(2)` is `ax+by=0 ` ...(i) The equation of the required circle is `{(x-a)^(2)+y^(2)-a^(2)} + lambda(ax+by)=0` Since (i) is a diameter of circle (ii). `:. a{-((a lambda-2a))/(2)}+b (-(lambda b)/(2))=0 rArr lambda = (2^(2))/(a^(2)+b^(2))` Putting the value of `lambda` in (ii), the equation of the required circle is `(a^(2)+b^(2))(x^(2)+y^(2))=2ab(bx-ay)` |
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