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Examples on Tangents to Circle Set 3

Examples On Tangents To Circles Set-3

Example - 23

DIRECTOR CIRCLE

A point $P$ moves in such a way so that the tangents drawn from it to the circle ${x^2} + {y^2} = {a^2}$ are perpendicular. Find the locus of $P$.

Solution: To contrast between the various alternatives available to us, we will use all of them here:

From the figure, it is apparent that $OTPS$ is a square since all the angles are right angles and $OS = OT = a$ .

Thus, $\,OP = \sqrt 2 \,a$ , i.e., the distance of $P$ from $O$ is always $\sqrt 2 \,a,$ i.e., $P$ lies on a circle of radius $\sqrt 2 \,a$. Thus, $P$ satisfies the equation

\[\fbox{$\begin{array}{*{20}{c}} {{x^2} + {y^2} = 2{a^2}}\end{array}$}\]

This circle is called the Director circle of the given circle.

A CO-ORDINATE APPROACH - I

Let $P$ be the point $(h,k).$ The equation of the pair of tangents drawn from $P$ to the circle is

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{T^2} = S{S_1}\\ \Rightarrow \qquad &{(hx + ky - {a^2})^2} = ({x^2} + {y^2} - {a^2})({h^2} + {k^2} - {a^2})\end{align}\]

This combined equation will represent a pair of perpendicular straight lines if

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,&{\rm{Coeff}}{\rm{. of }}{x^2} + {\rm{Coeff}}{\rm{. of }}{y^2} = 0\\ \Rightarrow \qquad &({k^2} - {a^2}) + ({h^2} - {a^2}) = 0\\ \Rightarrow \qquad & {h^2} + {k^2} = 2{a^2}\end{align}\]

Using $(x,y)$ instead of $(h,k)$, we obtain the locus of $P$ in conventional form:

\[\begin{align}{{x^2} + {y^2} = 2{a^2}}\end{align}\]

A CO-ORDINATE APPROACH - II

The equation of any tangent to the circle ${x^2} + {y^2} = {a^2}$ can be written as $y = mx + a\sqrt {1 + {m^2}} .$

If this line passes through $P(h,k),$ the co-ordinates of $P$ must satisfy this equation:

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&k = mh + a\sqrt {1 + {m^2}} \\\\ \Rightarrow \qquad & {(k - mh)^2} = {a^2}(1 + {m^2})\\\\ \Rightarrow \qquad & ({h^2} - {a^2}){m^2} - 2mhk + ({k^2} - {a^2}) = 0\end{align}\]

This is a quadratic in $m$ and will yield two values, say ${m_1}$ and ${m_2},$ which physically corresponds to the fact that two tangents can be drawn from $P(h,k)$ to the circle.

The two tangents are at a right angle if ${m_1}{m_2} = - 1$

\[\begin{align}{} \Rightarrow\quad & \frac{{{k^2} - {a^2}}}{{{h^2} - {a^2}}} = - 1\\ \Rightarrow \quad & {h^2} + {k^2} = 2{a^2}\end{align}\]

Using $(x,y)$ instead of $k(h,k)$ we obtain the locus of $P$:

\[x^2 + y^2 = 2{a^2}\]

If this question were to be encountered in an exam, the pure-geometric approach would certainly turn out to be the fastest !

Example - 24

${C_1}$ and ${C_2}$ are two concentric circles, the radius of ${C_2}$ being twice that of ${C_1}$. From a point $P$ on ${C_2},$ tangents $PA$ and $PB$ are drawn to ${C_1}.$ Prove that the centroid of $\Delta PAB$ is on ${C_1}.$

Solution: Let us again attempt this question using both a pure-geometric and a co-ordinate approach.

Pure-geometric approach:

Recall the following straightforward theorem pertaining to right-angle triangles.

$\Delta$ $ABC$ is right-angled at $B$ and $BD$ is the median drawn from $B$ to the opposite side $AC$. Then our theorem tells us that

\[BD=AD=CD\]

This can be proved using simple geometry

We will put this theorem to use in the current example.

Since the radius $r_2 $ of $C_2$ is twice that of $C_1 $ ($r_1$ ) we have

\[\begin{align} &PO = 2\;OD\\ \Rightarrow \qquad & OD = PD & & & ...(1) \end{align}\]

Thus,$D$ is the mid-point of $PO$. This means that in $\Delta OAP$, $AD$ is the median to $OP$. By the theorem mentioned above, we have

\[AD = OD = PD = {r_1}\]

Thus, in quadrilateral $ADBO,$ we have $AD = BD = OA = OB = {r_1}.$ In other words, $ADBO$ is a parallelogram so $E$ is the midpoint of $OD.$ Thus, $\begin{align}ED =\frac {r_1}{2} = \frac{1}{2}OD\end{align}$ ...(2)

Also, since $AE = EB,$ $PE$ is the median of $\Delta PAB.$ Thus the centroid lies on $PE.$

From (1) and (2), we finally obtain $PD = 2ED.$ Thus, $D$ divides the median $PE$ in the ratio $2 : 1$ implying $D$ is the centroid which lies on ${C_1}.$

The descriptive nature of the pure-geometric solution just provided might make it appear to be very long but actually only a few simple elementary geometry facts have been used.

Co-ordinate approach

There’s no loss of generality in assuming that the two circles are centred at the origin. Thus, we can write their equations as

\[\begin{align} {{C_1}\,\,\,:}&{{x^2} + {y^2} = {r^2}}\\ {{C_2}\,\,\,:}&{{x^2} + {y^2} = 4{r^2}} \end{align}\]

Assume the point $P$ to have the co-ordinates $(h,k).$ The equation of $AB$ (the chord of contact) can then be written as

\[{\rm{Equation}}\,{\rm{of}}\;AB:\,\,\,hx + ky = {a^2}\]

We can evaluate the co-ordinates of $A({x_1},{y_1})$ and $B({x_2},{y_2})$ by simultaneously solving the equations for ${C_1}$ and $AB$.

Thus, ${x_1}$ and ${x_2}$ will be the roots of

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{x^2} + {\left( {\frac{{{r^2} - hx}}{k}} \right)^2} = {r^2}\\\\ \Rightarrow \qquad &({h^2} + {k^2}){x^2} - 2{r^2}hx + {r^2}({r^2} - {k^2}) = 0\\\Rightarrow \qquad & {x_1} + {x_2} = \frac{{2{r^2}h}}{{{h^2} + {k^2}}}\;\; and \;\;{x_1}{x_2} = \frac{{{r^2}({r^2} - {k^2})}}{{{h^2} + {k^2}}} \qquad \qquad ... (3) \end{align}\]

${y_1}$ and ${y_2}$ will be the roots of

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{\left( {\frac{{{r^2} - ky}}{h}} \right)^2} + {y^2} = {r^2}\\\\ \Rightarrow\qquad &({h^2} + {k^2}){y^2} - 2{r^2}ky + {r^2}({r^2} - {h^2}) = 0\\ \Rightarrow \qquad & {y_1} + {y_2} = \frac{{2{r^2}k}}{{{h^2} + {k^2}}}\;\; and \;\;{y_1}{y_2} = \frac{{{r^2}({r^2} - {h^2})}}{{{h^2} + {k^2}}} \qquad \qquad ... (4)\end{align}\]

If we let $(t,s)$ be the co-ordinates of the centroid $G$ of $\Delta PAB,$ we have

\[\left. \begin{align}{l}t = \frac{{{x_1} + {x_2} + h}}{3} & & \Rightarrow & 3t = h\left( {1 + \frac{{2{r^2}}}{{{h^2} + {k^2}}}} \right)\\s = \frac{{{y_1} + {y_2} + k}}{3} & & \Rightarrow & 3s = k\left( {1 + \frac{{2{r^2}}}{{{h^2} + {k^2}}}} \right)\end{align} \right\}\,\,\,\,{\rm{Using}}\left( 3 \right){\rm{and}}\left( 4 \right)\]

Finally, observe that

\[\begin{align}9({t^2} + {s^2}) = ({h^2} + {k^2}){\left( {1 + \frac{{2{r^2}}}{{{h^2} + {k^2}}}} \right)^2}\end{align}\]

But since $(h,k)$ lies on ${C_2},$ we have ${h^2} + {k^2} = 4{r^2}.$

Thus,

\[\begin{align} \,\,\,\,\,\,\,\,\,&9\left( {{t^2} + {s^2}} \right) = \left( {4{r^2}} \right)\left( {\frac{9}{4}} \right) = 9{r^2}\\ \Rightarrow \qquad & {t^2} + {s^2} = {r^2} \end{align}\]

implying that $G(t,s)$ lies on ${C_1}.$

Since now you are in a good position to compare the two approaches, which one could you have rather chosen for this question, the pure-geometric one or the co-ordinate one!

Example - 25

Consider the circle ${x^2} + {y^2} = {a^2}.$ A chord of this circle is bisected at the point $P({x_1},{y_1}).$ What is the equation of this chord ?

Solution: Convince yourself that such a chord will be unique, since it must be perpendicular to the line joining the origin to $P({x_1},{y_1}),$ as is clear from the figure below:

The slope of $AB$ then becomes $\begin{align}\frac{{ - {x_1}}}{{{y_1}}}\end{align}$ so its equation can be written simply as

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&y - {y_1} = - \frac{{{x_1}}}{{{y_1}}}(x - {x_1})\\ \Rightarrow \qquad & x{x_1} + y{y_1} = x_1^{\,\,2} + y_1^{\,\,2}\end{align}\]

To make this equation look “better”, we subtract ${a^2}$ from both sides

\[x{x_1} + y{y_1} - {a^2} = x_1^{\,\,\,2} + y_1^{\,\,\,2} - {a^2}\]

so that it can now be written concisely as

\[\fbox{$\begin{array}{*{20}{c}} {T({x_1},{y_1}) = S({x_1},{y_1})} \end{array}$}\]

Of course, this is easily generalised to the case when the equation of the circle is in the general form ${x^2} + {y^2} + 2gx + 2fy + c = 0;$ the result obtained is the same. The next example discusses a good application of this concept.