Question

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
1. x² + y²)² = 4R² x²y²
2. (x² + y²)³ = 4R² x²y²
3. (x² + y²)(x + y) = R²xy
4. (x² + y²)² = 4Rx²y²

Answer 1

Correct Answer - Option 2 : (x² + y²)³ = 4R² x²y²

Let the foot of perpendicular be P(h, k).

Then, the slope of line \({\rm{OP}} = \frac{{\rm{k}}}{{\rm{h}}}\)

Line AB is perpendicular to line OP, so slope of line \(AB = - \frac{h}{k}\)

\({\rm{y}} - {\rm{k}} = - \frac{{\rm{h}}}{{\rm{k}}}\left( {{\rm{x}} - {\rm{h}}} \right)\)

⇒ hx + ky = h² + k²

\(\Rightarrow {\rm{\;}}\frac{{\rm{x}}}{{\left( {\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{h}} \right)}} + \frac{{\rm{y}}}{{\left( {\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{{\rm{k}}}} \right)}} = 1\)

So, point A\(\left( {\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{{\rm{h}}},0} \right){\rm{and\;B}}\left( {0,\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{{\rm{k}}}} \right)\)

∵ ΔAOB is a right angled triangle, so AB is one of the diameter of the circle having radius R (given).

⇒ AB = 2R

\(\Rightarrow \sqrt {{{\left( {\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{{\rm{h}}}} \right)}^2} + {{\left( {\frac{{{{\rm{h}}^2} + {{\rm{k}}^2}}}{{\rm{k}}}} \right)}^2}} = 2{\rm{R}}\)

\(\Rightarrow {\rm{\;}}{\left( {{{\rm{h}}^2} + {{\rm{k}}^2}} \right)^2}\left( {\frac{1}{{{{\rm{h}}^2}}} + \frac{1}{{{{\rm{k}}^2}}}} \right) = 4{{\rm{R}}^2}\)

⇒ (h² + k²)³ = 4R² h²k²

On replacing h by x and k by y, we get

(x², x²)³ = 4R²x²y²

which is the required locus.