# Jee Examples on Equations of Circles Set-3

**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

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

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