Jee Examples On Equations Of Circles Set-2

Example -10

A fixed line \(L_1\)  intersects the co-ordinate axes at \(P(a, 0)\) and \(Q (0, b)\). A variable line \(L_2\), perpendicular to \(L_1\), intersects the axes at \(R\) and \(S\). Show that the locus of the points of intersection of \(PS\) and \(QR\) is a circle.

Solution: The equation of \(L_1\), using intercept form, can be written as

\[\begin{align}  &\frac{x}{a} + \frac{y}{b} = 1\\    \Rightarrow \qquad  & bx + ay = ab   \end{align}\]

Since \(L_2\) is perpendicular to \(L_1\), its equation can be written as

\[{L_2} \equiv ax - by + \lambda  = 0\]

where \(\begin{align}\lambda\end{align} \) is a real parameter.

Using the equation of \(L_2\), we can determine \(R\) and \(S\) to be \(\begin{align}\left( { - \frac{\lambda }{a},0} \right)\;{\rm{and}}\;\left( {0,\frac{\lambda }{b}} \right)\end{align}\) respectively.

We now write the equations to \(PS\) and \(QR\) using the two-point form:

\[\begin{align}   &PS:\frac{y}{{x - a}} = \frac{{ - \lambda }}{{ab}} \qquad \quad  \Rightarrow \qquad \quad \lambda x + aby = a\lambda  \qquad \qquad \qquad  ...(1)\\   &QR:\frac{{y - b}}{x} = \frac{{ab}}{\lambda } \qquad \quad \;\,  \Rightarrow  \qquad \quad  - abx + \lambda y = b\lambda  \qquad \qquad \quad \;  ...(2)   \end{align}\]

The relation that the intersection point of \(PS\) and \(QR\), will satisfy can be evaluated by eliminating \(\begin{align}\lambda \end{align}\) from (1) and (2). We thus obtain

\[\begin{align}   & \lambda  = \frac{{aby}}{{a - x}} = \frac{{abx}}{{y - b}}\\    \Rightarrow \qquad & ab{y^2} - a{b^2}y = {a^2}bx - ab{x^2}\\    \Rightarrow \qquad & {x^2} + {y^2} - ax - by = 0   \end{align}\]

This represents a circle centered at \(\begin{align}\left( {\frac{a}{2},\frac{b}{2}} \right)\end{align}\) and passing through the origin.

Example -11

Let \(\begin{align}\left( {{m_i},\frac{1}{{{m_i}}}} \right),\,i = 1,\,\,2,\,\,3,\,\,4\end{align}\) be four distinct points lying on a circle. Prove that \(\begin{align}{m_1}{m_2}{m_3}{m_4} = 1\end{align}\)

Solution: We first assume an equation for this circle \(C\), in its general form:

\[C:{x^2} + {y^2} + 2gx + 2fy + c = 0\]

Since \(\begin{align}\left( {{m_i},\frac{1}{{{m_i}}}} \right)\end{align}\) satisfies the equation of \(C\) for \(i = 1, 2, 3, 4\) we have

\[\begin{align}   & m_i^2 + \frac{1}{{m_i^2}} + 2g{m_i} + \frac{{2f}}{{{m_i}}} + c = 0 \qquad \qquad i = 1,\,2,\,3,\,4\\    \Rightarrow \qquad & m_i^4 + 2gm_i^3 + cm_i^2 + 2f{m_i} + 1 = 0 \qquad \quad \; i = 1,\,\,2,\,\,3,\,\,4   \end{align}\]

This last equation tells us that  \({{m}_{i}}s\)  are the roots of the following equation in \(m\):

\[\begin{align}{m^4} + 2g{m^3} + c{m^2} + 2fm + 1 = 0: & {\rm{Roots}}\;{\rm{of}}\;{\rm{this}}\;{\rm{equation}}\;{\rm{are}}\;m,\;i = 1,\;2,\;3,\;4\end{align}\]

The product of the roots, which is \(\begin{align}{m_1}{m_2}{m_3}{m_4}\end{align}\), can easily be seen to be \(1\) from this equation.

Example -12

Find the equation of the circle \(C\) which has two fixed points \(\begin{align}A\left( {{x_1},{y_1}} \right)\,\,{\rm{and}}\,B\left( {{x_2},{y_2}} \right)\,\end{align}\)  as the end-points of its diameter.

Solution: To evaluate the required equation, we can use a well known result from plane geometry: the angle in a semicircle is a right angle.

Thus, we can use this fact:

\[\begin{align}   (Slope\;of\;AP) \times &(Slope\;of\;PB) =  - 1\\    \Rightarrow \qquad & \frac{{y - {y_1}}}{{x - {x_1}}} \times \frac{{y - {y_2}}}{{x - {x_2}}} =  - 1\\    \Rightarrow \qquad & \left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0 &  & ...(1)   \end{align}\]

This is the required equation that will represent \(C\). Note that we could equivalently have used the Pythagoras theorem in \(\begin{align}\Delta APB\end{align}\) to evaluate the equation of \(C\):

\[\begin{align}   & A{P^2} + P{B^2} = A{B^2}\\    \Rightarrow \qquad & {\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} + {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} = {\left( {{x_1} - {x_2}} \right)^2} + {\left( {{y_1} - {y_2}} \right)^2}\\    \Rightarrow \qquad & 2{x^2} + 2{y^2} - 2x\left( {{x_1} + {x_2}} \right) - 2y\left( {{y_1} + {y_2}} \right) =  - 2{x_1}{x_2} - 2{y_1}{y_2}\\    \Rightarrow \qquad & \left( {x - {x_1}} \right)\left( {x - {x_2}} \right) + \left( {y - {y_1}} \right)\left( {y - {y_2}} \right) = 0   \end{align}\]

which is the same as what we obtained in (1).