JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If a circle of radius \(R\) passes through the origin \(O\) and intersects the coordinate axes at \(A\) and \(B,\) then the locus of the foot of perpendicular from \(O\) on \(AB\) is
- A \({({x^2} + {y^2})^2} = 4{R^2}{x^2}{y^2}\)
- B \({({x^2} + {y^2})^3} = 4{R^2}{x^2}{y^2}\)
- C \({({x^2} + {y^2})^2} = 4R{x^2}{y^2}\)
- D \(({x^2} + {y^2})(x + y) = {R^2}xy\)
Answer & Solution
Correct Answer
(B) \({({x^2} + {y^2})^3} = 4{R^2}{x^2}{y^2}\)
Step-by-step Solution
Detailed explanation
Slope of \(AB = \frac{{ - h}}{k}\) Equation of \(AB\) is \(hx + ky = {h^2} + {k^2}\) \(A\left( {\frac{{{h^2} + {k^2}}}{h},0} \right),B\left( {0,\frac{{{h^2} + {k^2}}}{k}} \right)\) \(As,AB = 2R\) \( \Rightarrow {\left( {{h^2} + {k^2}} \right)^3} = 4{R^2}{h^2}{k^2}\)…
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