Example –7

Let \({S_1} = 0\) and \({S_2} = 0\) be two circles with radii \({r_1}\)  and \({r_2}\) respectively. Find the locus of the point P at which the two circles subtend equal angles.

Solution: Consider  \({S_1} = 0\) which subtends an angle \(\alpha \) (say) at \(P(h,k)\)

We have,

\({C_1}T = {r_1}\) and \(PT = \sqrt {{S_1}(h,k)} \)

Thus, in \(\Delta {C_1}PT,\) we have

\[\tan \frac{\alpha }{2} = \frac{{{r_1}}}{{\sqrt {{S_1}(h,k)} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

We can write an exactly analogous equation for \({S_2}\) which subtends the same angle \(\alpha \) at P:

\[\tan \frac{\alpha }{2} = \frac{{{r_2}}}{{\sqrt {{S_2}(h,k)} }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\]

From (1) and (2), we have

\[\frac{{{r_1}}}{{\sqrt {{S_1}(h,k)} }} = \frac{{{r_2}}}{{\sqrt {{S_2}(h,k)} }}\]

Using \((x,y)\)  instead of \((h,k),\) we obtain the required locus as

\[\frac{{{S_1}(x,y)}}{{r_1^2}} = \frac{{{S_2}(x,y)}}{{r_2^2}}\]

Example –8

Circles are drawn passing through the origin O to intersect the co-ordinate axes at points P and Q such that \(m \cdot OP + n \cdot OQ\) is a constant. Show that the circles pass through a fixed point other than the origin.

Solution: Consider one such circle as shown in the figure below:

We have \(OP = p\) and \(OQ = q.\) Thus,

\[mp + nq = k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

Now, observe that the centre C of the circle will be \(\begin{align} \left( {\frac{p}{2},\frac{q}{2}} \right)\end{align}\)  and its radius will be OC. Thus, the equation of the circle is

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {x - \frac{p}{2}} \right)^2} + {\left( {y - \frac{q}{2}} \right)^2} = \frac{{{p^2}}}{4} + \frac{{{q^2}}}{4}\\&\Rightarrow  \quad {x^2} + {y^2} - px - qy = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)\end{align}\]

From (1), we can find q in terms of p and substitute in (2) so that (2) becomes

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^2} + {y^2} - px + \left( {\frac{{mp - k}}{n}} \right)y = 0\\&\Rightarrow \quad   \{ n({x^2} + {y^2}) - ky\}  + p\{  - nx + my\}  = 0\end{align}\]

We see that the variable circle can be written as

\[{S_0} + p{L_0} = 0,p \in \mathbb{R}\]

where \({S_0} \equiv n({x^2} + {y^2}) - ky = 0\) and \({L_0} \equiv  - nx + my = 0\). Thus, the variable circle will always pass through the two intersection points of \({S_0}\) and \({L_0}\), one of them obviously being the origin. The other fixed point can be obtained by simultaneously solving the equations of \({S_0}\) and \({L_0}\) to be

\[\left( {\frac{{mk}}{{{m^2} + {n^2}}},\frac{{nk}}{{{m^2} + {n^2}}}} \right)\]

Example –9

The two curves

\[\begin{array}{l}{S_1}:a{x^2} + 2hxy + b{y^2} - 2gx - 2fy + c = 0\\{S_2}:a'{x^2} - 2hxy + (a' + a - b){y^2} - 2g'x - 2f'y + c = 0\end{array}\]

intersect at four concyclic points and Let P be the point \(\begin{align}\left( {\frac{{g' + g}}{{a' + a}},\frac{{f' + f}}{{a' + a}}} \right).\end{align}\) Find the value of \(\begin{align}\frac{{P{A^2} + P{B^2} + P{C^2}}}{{P{D^2}}}.\end{align}\)

Solution: From what we’ve learnt in this chapter, we can write the equation of an arbitrary curve S passing through the intersection points of two given curves, here \({S_1}\) and \({S_2}\), as

\[S \equiv {S_1} + \lambda {S_2} = 0) \; where\; \lambda  \in \mathbb{R}\]

Here,

\[S:(a + \lambda a'){x^2} + 2h(1 - \lambda )xy + (b + \lambda (a' + a - b)){y^2} - 2x(g + g'\lambda ) - 2y(f + f') + c(1 + \lambda ) = 0\,...\left( 1 \right)\] Since this curve S passes through the four concyclic points \(A,\,\,B,\,\,C,\,\,D\) its equation in (1) must be that  of a circle. Thus, we impose the necessary constraints:

\[\begin{align}&{\text{Coeff}}{\text{. of }}\;{x^2} = {\text{Coeff}}{\text{. of }}\;{y^2} \quad  \Rightarrow  \quad  a + \lambda a' = b + \lambda (a' + a -b)\\&\qquad\qquad\qquad\qquad\qquad\qquad\;\;\Rightarrow \quad  \lambda  = 1\\&{\text{Coeff}}{\text{. of }}xy = 0 \qquad \qquad\quad\;\;  \Rightarrow  \quad  \lambda  = 1\end{align}\]

Thus, for \(\lambda  = 1,\) the equation in (1) represents a circle. This equation now becomes :

\[S:(a + a'){x^2} + (a + a'){y^2} - 2(g + g')x - 2(f + f')y + 2c = 0\]

The centre of S therefore is \(\begin{align}\left( {\frac{{g + g'}}{{a + a'}},\frac{{f + f'}}{{a + a'}}} \right).\end{align}\)  But this is the same as the point P!. Thus, P is the centre of the circle passing through \(A,\,\,B,\,\,C,\,\,D\)  so that

\[PA = PB = PC = PD\]

 Thus, we simply have

\[\begin{align}\frac{{P{A^2} + P{B^2} + P{C^2}}}{{P{D^2}}} = 3\end{align}\]

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