Jee Examples on Equations of Circles Set-3

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Example -13

Given the circle \(\begin{align}C:{x^2} + {y^2} + 2gx + 2fy + c = 0\end{align}\), find the intercepts that it makes on the \(x\)-axis and \(y\)-axis.

Solution: We did a similar case in Example - 8 by solving simultaneously the equation of the circle and the line on which the intercept is required. Here, we’ll proceed analogously.

x – intercept

From (1), we have \(\begin{align}{x_1} + {x_2} =  - 2g\,\,{\rm{and}}\,\,{x_1}{x_2} = c.\end{align}\)

Thus,

\[\begin{align}\left| {{x_1} - {x_2}} \right| &= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\&= 2\sqrt {{g^2} - c} \end{align}\]

y - intercept

Thus, the intercepts are of length \(\begin{align}2\sqrt {{g^2} - c} \,\,{\rm{and}}\,\,2\sqrt {{f^2} - c} \end{align}\) respectively. Obviously, if \(\begin{align}{g^2} < c\end{align}\) the circle and the x–axis do not touch/intersect; if  \({{f}^{2}}<c\)  the circle and the y–axis do not touch or intersect.

Example -14

What values can the variable a take so that the point \((a – 1, a + 1)\) lies inside the circle \(\begin{align}{x^2} + {y^2} - 12x + 12y - 62 = 0\end{align}\) but outside the circle \(\begin{align}{x^2} + {y^2} = 8\end{align}\).

Solution: We will use the general result that we derived in Example - 6 earlier (refer). Specifically, if \(S = 0 \) is the equation of circle and \(P (x_1, y_1)\) be any point, then

\[\begin{align}  &S\left( {{x_1},{y_1}} \right) < 0 \qquad \Rightarrow \qquad P\,\,{\text{lies inside}}\,S\\  &S\left( {{x_1},{y_1}} \right) > 0 \qquad \Rightarrow \qquad P\,\,{\text{lies outside}}\,S  \end{align}\]

Using these relations for the current case, we obtain and 

\[\begin{align}   {\left( {a - 1} \right)^2} + {\left( {a + 1} \right)^2} - &12\left( {a - 1} \right) + 12\left( {a + 1} \right) - 62 < 0 \qquad \qquad ...(1)\\   & \Rightarrow  \qquad 2{a^2} - 36 < 0\\   & \Rightarrow  \qquad  - 3\sqrt 2  < a < 3\sqrt 2  \qquad \qquad \quad \quad \;\; ...(i)\\   and\;{\left( {a - 1} \right)^2} +& {\left( {a + 1} \right)^2} - 8 > 0 \qquad \qquad \qquad \qquad \quad \quad \;\; ...(2)\\   & \Rightarrow  \qquad 2{a^2} - 6 > 0\\   & \Rightarrow  \qquad a > \sqrt 3 \,\,{\rm{or}}\,\,a <  - \sqrt 3  \qquad \qquad \quad ...(ii)   \end{align}\]

The intersection of (i) and (ii) gives us the required values of \(a\) as

\[a \in \left( { - 3\sqrt 2 , - \sqrt 3 } \right) \cup \left( {\sqrt 3 ,\,\,3\sqrt 2 } \right)\]

Example -15

Find the locus of the foot of the perpendicular drawn from the origin upon any chord of a circle \(\begin{align}S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0\end{align}\) which subtends a right angle at the origin.

Solution:    The situation is depicted graphically in the figure below to make things clearer:

Observe that the equation of the chord \(AB\) can be written as

\[\begin{align}  & \frac{{y - m}}{{x - l}} = \frac{{ - l}}{m} &  & \left( {\because AB \bot OX} \right) \hfill \\     \Rightarrow \qquad &  \; lx + my = {l^2} + {m^2} \hfill \\    \end{align} \]

Now, if we homogenize the equation of \(S\) using the equation of the chord \(AB\), what we’ll get is the equation of the pair of straight lines \(OA\) and \(OB\) (as discussed in the last chapter on straight lines). This is what we proceed to do:

\[\begin{align}{x^2} + {y^2} + 2gx\left( {\frac{{{l} x + my}}{{{l^2} + {m^2}}}} \right) + 2fy\left( {\frac{{lx + my}}{{{l^2} + {m^2}}}} \right) + c{\left( {\frac{{lx + my}}{{{l^2} + {m^2}}}} \right)^2} = 0 \qquad \qquad \qquad ...(1)\end{align}\]

This is the joint equation of \(OA\) and \(OB\), since \(OA\) and \(OB\) need to be at right angles, we impose the appropriate constraint for perpendicularity on (1):

\[\begin{align}   &{\rm{Coeff}}{\rm{. of }}{x^2} + {\rm{coeff}}.\,{\rm{of}}\,{y^2} = 0\\   &\left\{ {1 + \frac{{2gl}}{{{l^2} + {m^2}}} + \frac{{c{l^2}}}{{{{\left( {{l^2} + {m^2}} \right)}^2}}}} \right\}\,\, + \,\,\left\{ {1 + \frac{{2fm}}{{{l^2} + {m^2}}} + \frac{{c{m^2}}}{{{{\left( {{l^2} + {m^2}} \right)}^2}}}} \right\} = 0\\    \Rightarrow \qquad & 2 + \frac{{2gl}}{{{l^2} + {m^2}}} + \frac{{2fm}}{{{l^2} + {m^2}}} + \frac{{c\left( {{l^2} + {m^2}} \right)}}{{{{\left( {{l^2} + {m^2}} \right)}^2}}} = 0\\    \Rightarrow \qquad & {l^2} + {m^2} + gl + fm + \frac{c}{2} = 0   \end{align}\]

This is the equation of a circle. To be more conventional, we should use \((x, y)\) instead of the variables \(l\) and \(m\).

Thus, the required locus is

\[\begin{align}{x^2} + {y^2} + gx + fy + \frac{c}{2} = 0\end{align}\]

TRY YOURSELF - I

Q. 1    Find the greatest and least distances of the point \(P (10, 7)\) from the circle \(\begin{align}{x^2} + {y^2} - 4x - 2y - 20 = 0\end{align}\)

Q. 2  Through the origin \(O\), a straight line is drawn to cut the line \(\begin{align}y = mx + c\end{align}\) at \(P\). If \(Q\) is a point on this line such that \(\begin{align}OP \cdot OQ = {\lambda ^2}\end{align}\), show that the locus of \(Q\) is a circle passing through the origin.

Q. 3    What is the area of an equilateral triangle inscribed in the circle \(\begin{align}{x^2} + {y^2} + 2gx + 2fy + c = 0?\end{align}\)

Q. 4    Determine the equation of the circle passing through the points \((1, 2)\) and \((3, 4)\) and touching the line \(\begin{align}3x + y - 3 = 0.\end{align}\)

Q. 5    A circle whose centre is the point of intersection of the lines \(\begin{align}2x - 3y + 4 = 0\end{align}\) and \(\begin{align}3x + 4y - 5 = 0\end{align}\) passes through the origin. Find its equation.

Q. 6    Find the equation of the circle which touches the \(x\)-axis and two of whose diameters lie along \(\begin{align}2x - y - 5 = 0\,\,{\rm{and}}\,\,3x - 2y - 8 = 0\end{align}\)

Q. 7    Find the equation of the circumcircle of an equilateral triangle two of whose vertices are \((–1, 0)\) and \((1, 0)\) and the third vertex lies above the \(x\)–axis.

Q. 8    Find the equation of the circle passing through \((1,0)\) and \((0,1)\) and having the smallest possible radius.

Q. 9    The equations of the sides of a quadrilateral are given by  \(\begin{align}{L_r} = {a_r}x + {b_r}y + {c_r} = 0\,\,\left( {r = 1,2,3,4} \right).\end{align}\) If the quadrilateral is concyclic, show that

\[\frac{{{a}_{1}}{{a}_{3}}-{{b}_{1}}{{b}_{3}}}{{{a}_{2}}{{a}_{4}}-{{b}_{2}}{{b}_{4}}}=\frac{{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}}}{{{a}_{2}}{{b}_{4}}+{{a}_{4}}{{b}_{2}}}\]

Q. 10    Find the point on the straight line \(\begin{align}y = 2x + 11\end{align}\) which is nearest to the circle

\[\begin{align}16\left( {{x^2} + {y^2}} \right) + 32x - 8y - 50 = 0\end{align}\]

 

Download SOLVED Practice Questions of Jee Examples on Equations of Circles Set-3 for FREE
Circles
grade 11 | Questions Set 1
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Download SOLVED Practice Questions of Jee Examples on Equations of Circles Set-3 for FREE
Circles
grade 11 | Questions Set 1
Circles
grade 11 | Answers Set 1
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grade 11 | Questions Set 2
Circles
grade 11 | Answers Set 2
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