# Circles Set-5

**Example –13**

A circle of radius *r* passes through the origin *O* and cuts the axes at *A* and *B*. Show that the locus of the foot of the perpendicular from *O* to *AB* is

\[{({x^2} + {y^2})^2}\left( {\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}}} \right) = 4{r^2}\]

**Solution: **Let the co-ordinates of *A* and *B* be \((a,0)\) and \((0,b)\) respectively, so that (like in Example - 8), the equation to the variable circle becomes

\[{x^2} + {y^2} - ax - by = 0\]

We have,

\[{a^2} + {b^2} = 4{r^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

Let the foot of perpendicular *P* have the co-ordinates \((h,k).\) Since \(OP \bot AB,\) we obtain

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{k}{h} \times \frac{b}{{ - a}} = - 1\,.\\&\Rightarrow \quad \frac{k}{a} = \frac{h}{b} = \frac{{\sqrt {{h^2} + {k^2}} }}{{\sqrt {{a^2} + {b^2}} }} = \frac{{\sqrt {{h^2} + {k^2}} }}{{2r}}\\&\Rightarrow \quad a = \frac{{2rk}}{{\sqrt {{h^2} + {k^2}} }},\,\,b = \frac{{2rh}}{{\sqrt {{h^2} + {k^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{align}\]

Using (2) in (1) and \((x,y)\) instead of \((h,k),\) we obtain the required locus.

**Example –14**

Find the condition so that the chord \(x\cos \alpha + y\sin \alpha = p\) subtends a right angle at the centre of the circle \({x^2} + {y^2} = {a^2}\).

**Solution: **

We can obtain the joint equation *J* to the pair of lines *OA *and *OB *by homogenizing the equation of the circle using the equation of the chord *AB*.

\[J:{x^2} + {y^2} = {a^2}{\left( {\frac{{x\cos \alpha + y\sin \alpha }}{p}} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

For *J* to represent two perpendicular lines, we must have in (1),

\[{\rm{Coeff}}{\rm{. of }}\;{x^2} + {\rm{Coeff.}}\;{\rm{of }}\;{y^2} = 0\]

which upon simplification yields the required condition as

\[{a^2} = 2{p^2}\]

**Example –15**

Suppose that the lines \({a_1}x + {b_1}y + {c_1} = 0\) and \({a_2}x + {b_2}y + {c_2} = 0\) intersect the co-ordinate axes at points \(A,\,\,B\,\,\) and \(C,\,\,D\) respectively. Find the condition that must be satisfied if these four points are to be concyclic.

**Solution: **The co-ordinate of \(A,\,\,B\,\,\) and \(C,\,\,D\) can be evaluated to be \(\begin{align}\left( { - \frac{{{c_1}}}{{{a_1}}},0} \right),\,\,\left( {0, - \frac{{{c_1}}}{{{b_1}}}} \right),\,\,\left( { - \frac{{{c_2}}}{{{a_2}}},0} \right)\end{align}\) and \(\begin{align}\left( {0, - \frac{{{c_2}}}{{{b_2}}}} \right).\end{align}\)

Instead of resorting to a detailed calculation, we simply use the result on tangents and secants that we’ve already derived earlier:

\(OA \cdot OC = OD \cdot OB = {l^2}\)

where \(l\) is the length of the tangent drawn from *O* to the circle. This gives

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - \frac{{{c_1}}}{{{a_1}}}} \right) \cdot \left( { - \frac{{{c_2}}}{{{a_2}}}} \right) = \left( { - \frac{{{c_2}}}{{{b_2}}}} \right) \cdot \left( { - \frac{{{c_1}}}{{{b_1}}}} \right)\\&\Rightarrow\quad {a_1}{a_2} - {b_1}{b_2} = 0\end{align}\]

This is the required condition!

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