Examples On Tangents To Circles Set-4

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Example - 26

Find the locus of the mid-point of the chords of the circle \({x^2} + {y^2} = {a^2}\) which subtend a right angle at the centre.

Solution:

Since \(AB\) is bisected at \(M(h,k),\) we can use the result obtained in the last example to write the equation of \(AB\):\[\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&T(h,k) = S(h,k)\\ \Rightarrow  & hx + ky - {a^2} = {h^2} + {k^2} - {a^2}\\ \Rightarrow  & hx + ky = {h^2} + {k^2}\end{array}\]

We can view the chord \(AB\) as a line intersecting the curve (circle) \({x^2} + {y^2} = {a^2}.\)  Thus, we can obtain the joint equation of \(OA\) and \(OB\) by homogenizing the equation of the circle using the equation of the chord AB:

\[\begin{align}{x^2} + {y^2} - {a^2}{\left( {\frac{{hx + ky}}{{{h^2} + {k^2}}}} \right)^2} = 0\end{align}\]

Joint equation of

OA and OB:

\[ \Rightarrow \qquad {({h^2} + {k^2})^2}({x^2} + {y^2}) - {a^2}{(hx + ky)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

Since \(OA\) and \(OB\) are perpendicular, we must have

\[{\text{Coeff }}{\text{.of }}{x^2} + {\text{Coeff.}}{\text{ of }}{y^2} = 0\]

\[ \Rightarrow \qquad 2{({h^2} + {k^2})^2} - {a^2}{h^2} - {a^2}{k^2} = 0\]

\[ \Rightarrow \qquad 2{({h^2} + {k^2})^2} - {a^2}{h^2} - {a^2}{k^2} = 0\]

Using \((x,y)\) instead of\((h,k),\) we obtain the following equation as the locus of \(M\):

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

This is a circle concentric with the original circle as might have been expected.

Try solving this question using pure geometric considerations; the solution will be much simpler.

Example - 27

Consider two circles with the following equations:

\[\begin{array}{l}{C_1}:\,\,\,\,\,{x^2} + {y^2} - 2x - 2y + 1 = 0\\{C_2}:\,\,\,\,\,{x^2} + {y^2} - 16x - 2y + 61 = 0\end{array}\]

Find the values that a can take so that the variable line \(y = 2x + a\) lies between these two circles without touching or intersecting either of them.

Solution: Observe carefully that what is variable about the variable line \(y = 2x + a\) is not its slope but its   \(y\)-intercept \(a\). Thus, we can always adjust \(a\) so that line stays between the two circles. The following diagram makes this clear.

Evaluate \(\;{a_{\max }}\) : The line \(y = 2x + {a_{\max }}\) is a tangent to \({C_1}\) if the perpendicular distance of the centre \((1, 1)\) of \({C_1}\)  from this line is equal to \(C_1^{\,'}s\) radius which is \(1\). Thus :

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{{\left| {2 - 1 + {a_{\max }}} \right|}}{{\sqrt 5 }} = 1\\ \Rightarrow \qquad & 1 + {a_{\max }} =  \pm \sqrt 5 \\ \Rightarrow \qquad  & {a_{\max }} =  - \sqrt 5  - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{\text{since from the figure we }}\\{\text{can see that }}{a_{\max }}{\text{is definitely}}\\{\text{negative}}\end{array} \right\}\end{align}\]

Evaluate \({a_{\min }}\) :   The distance of \(C_2^{\,'}s\) center \((8, 1)\) from \(y = 2x + {a_{\min }}\) must be equal to its radius which is equal to \(2\). Thus :

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{{\left| {16 - 1 + {a_{\min }}} \right|}}{{\sqrt 5 }} = 2\\ \Rightarrow  & 15 + {a_{\min }} =  \pm 2\sqrt 5 \\ \Rightarrow  & {a_{\min }} = 2\sqrt 5  - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{\text{we have selected the larger of the two}}\\{\text{values possible since that is what }}\\{\text{corresponds to }}{a_{\min }},{\text{i}}{\rm{.e}}{\text{. because the}}\\{\text{line }}y = 2x + {a_{\min }}{\text{ lies above }}{\text{ }}{C_2}.\end{array} \right\}\end{align}\]

Thus, we obtain the allowed values of a as

\[2\sqrt 5  - 15 < a <  - \sqrt 5  - 1\]

Example - 28

A circle touches the line \(y = x\) at a point \(P\) such that \(OP = 4\sqrt 2 \) where \(O\) is the origin. The circle contains the point \((–10, 2)\) in its interior and the length of its chord on the line \(x + y = 0\) is \(6\sqrt 2 .\) Determine the equation of the circle.

Solution: As always, before starting with the solution, it is a good to draw a diagram of the situation described to get a feel of it. Also, as far as possible, we should try to use pure-geometric considerations to cut down on the (complicated) algebraic manipulations that would result otherwise.

We have,

\[\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&PX \bot (y = x)\\ \Rightarrow \qquad & \frac{{k + 4}}{{h + 4}} =  - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\\ \Rightarrow \qquad & h + k + 8 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(2)\end{align}\]

Also,

\[\begin{align}&\left\{ \begin{array}{l}{\rm{Perpendicular dist}}{\rm{.}}\\{\rm{of }}X{\rm{ from }}x + y = 0\end{array} \right\} = BX = \sqrt {A{X^2} - A{B^2}}  = \sqrt {{r^2} - A{B^2}} \\ \Rightarrow \qquad & \frac{{\left| {h + k} \right|}}{{\sqrt 2 }} = \sqrt {{{(h + 4)}^2} + {{(k + 4)}^2} - {{(3\sqrt 2 )}^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(3)\end{align}\]

Using (1) and (2) in (3), we obtain

\[\begin{align}&{\left( {4\sqrt 2 } \right)^2} = 2{\left( {h + 4} \right)^2} - 18\\ \Rightarrow \qquad  & h + 4 =  \pm 5\\ \Rightarrow \qquad & h =  - 9,1\end{align}\]

Given the region in which \(X\) lies, \(h\) must be \(–9\). Thus, from (2), \(k\) is \(1\) and the radius \(r\) is \(5\sqrt 2 \) .

The required equation is therefore

\[\begin{align}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{(x + 9)^2} + {(y - 1)^2} = {(5\sqrt 2 )^2}\\ \Rightarrow \qquad & {x^2} + {y^2} + 18x - 2y + 32 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(4)\end{align}\]

For students used to rigor, it can finally be verified that the circle given by (4) does indeed contain the point \((-10, 2)\) and thus our initial assumption of the region in which the centre \(X\) lies, was correct.

 

Download SOLVED Practice Questions of Examples On Tangents To Circles Set-4 for FREE
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Download SOLVED Practice Questions of Examples On Tangents To Circles Set-4 for FREE
Circles
grade 11 | Questions Set 1
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grade 11 | Questions Set 2
Circles
grade 11 | Answers Set 2
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