Examples On Tangents To Circles Set-3

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Example - 23                                             


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.


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:

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


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


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 \({h^2} + {k^2} = 4{r^2}.\)


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

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.


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