Examples On Tangents To Circles Set-3

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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.