Example – 1

The circle \({x^2} + {y^2} - 4x - 4y + 4 = 0\) is inscribed in a triangle which has two of its sides along the co-ordinate axes. If the locus of the circumcentre of the triangle is

\[x + y - xy + k\sqrt {{x^2} + {y^2}}  = 0\]

find the value of k.

Solution: The situation is described clearly in the figure below:

The equation of L is, using the intercept form,

\[L:\frac{x}{a} + \frac{y}{b} = 1\]

The distance of the centre of S, i.e. (2, 2) from L must equal the radius of S which is 2. Thus,

\[\frac{{\left| {\frac{2}{a} + \frac{2}{b} - 1} \right|}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{align}&{\text{We now use the fact that }}\\&L{\text{(2, 2) is negative since}}\\&{\text{(2, 2) and the origin lie }}\\&{\text{on the same side of }}L{\text{ and}}\\&L{\text{(0, 0) is negative }}\end{align} \right\}\]

\[ \Rightarrow \quad  2a + 2b - ab + 2\sqrt {{a^2} + {b^2}}  = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)\]

From pure geometric considerations, the circumcentre C of \(\Delta OAB\) lies on AB and is in fact, the  mid-point of AB.

Thus,

\[C \equiv \left( {\frac{a}{2},\frac{b}{2}} \right)\]

Slightly manipulating (1), we obtain

\[\frac{a}{2} + \frac{b}{2} - \left( {\frac{a}{2}} \right)\left( {\frac{b}{2}} \right) + \sqrt {{{\left( {\frac{a}{2}} \right)}^2} + {{\left( {\frac{b}{2}} \right)}^2}}  = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\]

The locus of \(\begin{align}\left( {\frac{a}{2},\frac{b}{2}} \right)\end{align}\) is given by (2). Using \((x,y)\) instead of \(\begin{align}\left( {\frac{a}{2},\frac{b}{2}}\right),\end{align}\) we obtain

\[x + y - xy + \sqrt {{x^2} + {y^2}}  = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(3)\]

Upon comparing (3) with the locus specified in the question, we obtain \(k = 1.\)

Example – 2

Consider a curve \(a{x^2} + 2hxy + b{y^2} = 1\) and a point P not on the curve. A line drawn from P intersects the curve at points Q and R. If the product \(PQ \cdot PR\) is independent of the slope of the line, then show that the curve is a circle.

Solution: Since distances are involved from a fixed point, it would be a good idea to use the polar form of the line to write the co-ordinates of Q and R.

Let P be \(({x_1},{y_1})\) and let \(\theta \) denote the slope of the variable line. For any point \((x,y)\)  lying on this line at a distance r from P, we have

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{x - {x_1}}}{{\cos \theta }} = \frac{{y - {y_1}}}{{\sin \theta }} = r\\&\quad\Rightarrow  \quad x = {x_1} + r\cos \theta \,\,;\,\,y = {y_1} + r\sin \theta \end{align}\]

If \((x,y)\)  lies on the given curve, it must satisfy the equation of the curve :

\[\begin{align}&a{({x_1} + r\cos \theta )^2} + b{({y_1} + r\sin \theta )^2} + 2h({x_1} + r\cos \theta )({y_1} + r\sin \theta ) = 1\\& \Rightarrow \quad  \{ a{\cos ^2}\theta  + b{\sin ^2}\theta  + h\sin 2\theta \} {r^2} + 2\{ a{x_1}\cos \theta  + b{y_1}\sin \theta  + h({x_1}\cos \theta \\&\qquad\qquad \qquad\qquad + {y_1}\sin \theta )\} r + ax_1^2 + by_1^2 + 2h{x_1}{y_1} - 1 = 0\qquad\qquad \quad...(1)\end{align}\]

This quadratic has two roots in r, say \({r_1}\) and \({r_2}\), which will actually correspond to \(PQ\) and \(PR\) since Q and R lie on the curve. Thus, \(PQ \cdot PR\) being independent of \({\rm{\theta }}\) means that \({r_1}{r_2}\) for (1) is independent of \({\rm{\theta }}\) i.e.

\[\begin{align} & \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{a{{\cos }^{2}}\text{ }\!\!\theta\!\!\text{ }+b{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ }+h\sin 2\text{ }\!\!\theta\!\!\text{ }}\,\,\text{is independent of}\,\,\theta  \\ \\
 & \Rightarrow \quad \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{\left( \frac{a+b}{2} \right)+\left( \frac{a-b}{2} \right)\cos 2\text{ }\!\!\theta\!\!\text{ }+h\sin 2\text{ }\!\!\theta\!\!\text{ }}\,\,\text{is independent of}\,\,\theta  \\ \\
 & \left\{ \text{The denominator was obtained in this form by writing }{{\cos }^{2}}\theta \text{ as }\frac{1+\cos 2\theta }{2}\text{ and }{{\sin }^{2}}\theta \text{ as }\frac{1-\cos 2\theta }{2} \right\} \\ 
 & \Rightarrow \quad  \frac{ax_{1}^{2}+by_{1}^{2}+2h{{x}_{1}}{{y}_{1}}-1}{\left( \frac{a+b}{2} \right)+\sqrt{{{\left( \frac{a-b}{2} \right)}^{2}}+{{h}^{2}}}\{\sin (2\theta +\phi \}}\,\,\text{is independent of}\,\,\theta  \\ \\
 & \left\{ \text{The denominator was obtained by combining the two trignometric terms}\text{.}\tan \phi \,\,\text{is }\left( \frac{a-b}{2h} \right)\text{ } \right\} \\ 
 & \Rightarrow \quad \text{This is only possible  when}{{\left( \frac{a-b}{2} \right)}^{2}}+{{h}^{2}}=0 \\ 
 & \Rightarrow \quad a=b\,\,\,and\,\,\,h=0 \\ 
\end{align}\]

Thus, the equation of the given curve reduces to

\(a{x^2} + a{y^2} = 1\)

which is clearly the equation of a circle.

Example – 3

The equation of a circle is \(S:2x(x - a) + y(2y - b) = 0\) where \(a,\,\,b \ne 0.\)  Find the condition on a and b if two chords each bisected by the x-axis, can be drawn to the circle from the point \(\begin{align}P\left( {a,\frac{b}{2}} \right).\end{align}\)

Solution: Observe that the centre of S is \(\begin{align}\left( {\frac{a}{2},\frac{b}{4}} \right)\end{align}\) and P lies on the circle:

Let us consider a chord \(PQ\) of the circle bisected at the x-axis, say, at the point \((h,0).\) We can write the equation of \(PQ\) as :

\[T(h,0) = S(h,0)\]

where \(S(x,y)\) is

\[\begin{align}   &{x^2} + {y^2} - ax - \frac{b}{2}y = 0\\    \Rightarrow  \qquad & hx - \frac{a}{2}(x + h) - \frac{b}{4}(y + 0) = {h^2} - ah\\    \Rightarrow \qquad & \left( {h - \frac{a}{2}} \right)x - \frac{b}{4}y + \left( {\frac{{ah}}{2} - {h^2}} \right) = 0 \qquad\qquad ...(1)    \end{align}\]

Since this chord passes through \(\begin{align}P\left( {a,\frac{b}{2}} \right),\end{align}\)  the co-ordinates of (P) must satisfy (1). Thus,

\[\begin{align}&\qquad\; \left( {h - \frac{a}{2}} \right)a - \frac{b}{4}\left( {\frac{b}{2}} \right) + \frac{{ah}}{2} - {h^2} = 0\\& \Rightarrow  \quad \frac{{3ah}}{2} - \frac{{{a^2}}}{2} - \frac{{{b^2}}}{8} - {h^2} = 0\\& \Rightarrow  \quad {h^2} - \frac{{3ah}}{2} + \frac{{{a^2}}}{2} + \frac{{{b^2}}}{8} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{align}\]

We want to have two such possible chords PQ bisected at the x-axis. Thus, we must have two distinct values of h which can happen if the discriminant of (2) is positive. Thus,

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{9{a^2}}}{4} > 4\left( {\frac{{{a^2}}}{2} + \frac{{{b^2}}}{8}} \right)\\ &\Rightarrow \qquad   {a^2} > 2{b^2}\end{align}\]

This is the required condition that a and b must satisfy.

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