Examples On Common Chords And Radical Axes Set-2

Example - 34

Prove that two circles, both of which pass through the point \((0,a)\) and \((0, - a)\) and touch the line \(y = mx + c,\) will cut orthogonally if \({c^2} = {a^2}(2 + {m^2})\)

Solution: It should be clear that the \(y\)-axis is the common chord of the two circles \({C_1}\) and \({C_2}\) :

Let the centres of \({C_1}\) and \({C_2}\) be  \(( - {g_1},0)\) and \(( - {g_2},0)\) so that their radii become \(\sqrt {g_1^2 + {a^2}} \) and \(\sqrt {g_2^2 + {a^2}} \) respectively. Their equations then become:

\[{C_1}:{x^2} + {y^2} + 2{g_1}x - {a^2} = 0\]

\[{C_2}:{x^2} + {y^2} + 2{g_2}x - {a^2} = 0\]

The line \(y = mx + c\) touches both \({C_1}\) and \({C_2}\,\) so that the perpendicular distance of the centres of \({C_1}\) and \({C_2}\,\) from this line must be respectively equal to their radii. Thus, we obtain

\[\begin{align}&\frac{{\left| {m{g_1} - c} \right|}}{{\sqrt {1 + {m^2}} }} = \sqrt {g_1^2 + {a^2}}\;\text{and} \;\frac{{\left| {m{g_2} - c} \right|}}{{1 + {m^2}}} = \sqrt {g_2^2 + {a^2}} \\
\Rightarrow \qquad\;\; &g_1^2 + 2mc{g_1} + {a^2}(1 + {m^2}) - {c^2} = 0\\
\;\text{and} \qquad &g_2^2 + 2mc{g_2} + {a^2}(1 + {m^2}) - {c^2} = 0\end{align}\]

Thus, \({g_1}\) and \({g_2}\) are the roots of the equation

\[{g^2} + 2mcg + {a^2}(1 + {m^2}) - {c^2} = 0\]

so that

\[{g_1}{g_2} = {a^2}(1 + {m^2}) - {c^2}\]

Finally, \({C_1}\) and \({C_2}\) are orthogonal if the condition

\[2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}\]

is satisfied, i.e.

\[\begin{align}  & 2\{ ({a^2}(1 + {m^2}) - {c^2}) + (0)(0)\}  =  - 2{a^2}\\     \Rightarrow \qquad & {c^2} = {a^2}(2 + {m^2})   \end{align}\]

This is the required relation for  and \({C_2}\) to be orthogonal.

Example - 35

Consider the following two circles:

\[\begin{align} &{S_1}:{x^2} + {y^2} - 16 = 0\\  &{S_2}:{x^2} + {y^2} - 8x - 12y + 16 = 0\end{align}\]

Circles are drawn which are orthogonal to both these circles. Tangents are drawn from the centre of the variable circle to \({S_1}\). Find the mid locus of the mid-point of the chord of contact so formed.

Solution: The centre of the variable circle, say \(({x_1},{y_1}),\) lies on the radical axis of the two given circles, i.e. on \({S_1} - {S_2} = 0\) by the result obtained in Example - 33.

Thus, \(({x_1},{y_1})\) must satisfy

\[\begin{align}   & {S_1} - {S_2} = 0\;{\rm{or}}\;8x + 12y - 32 = 0\\     \Rightarrow \qquad & 8{x_1} + 12{y_1} - 32 = 0 &  & ...(1)   \end{align}\]

From \(({x_1},{y_1}),\) two tangents are drawn to \({S_1}.\) The equation of the chord of contact is therefore

\[\begin{align}  &T({x_1},{y_1}) = 0\\     \Rightarrow  \qquad & x{x_1} + y{y_1} = 16 \qquad\qquad\qquad ...(2)   \end{align}\]

Let the mid-point of the chord of contact so formed be \(M(h,k).\) We can also write the equation of the same chord of contact using the equation we derived for a chord bisected at a given point.

Thus, the chord of contact can also be represented by the equation

\[\begin{align}& T(h,k) = S(h,k)\\\Rightarrow \qquad & hx + ky = {h^2} + {k^2} \qquad \qquad ...(3)\end{align}\]

Since (2) and (3) are the same lines, we have

\[\begin{align} & \frac{{{x_1}}}{h} = \frac{{{y_1}}}{k} = \frac{{16}}{{{h^2} + {k^2}}}\\ \Rightarrow \qquad & {x_1} = \frac{{16h}}{{{h^2} + {k^2}}};\,\,{y_1} = \frac{{16k}}{{{h^2} + {k^2}}}\qquad \qquad \dots \rm{(4)}\end{align}\]

Using (4) in (1), we finally obtain a relation in \((h,k)\):

\[\begin{align} & 8{x_1} + 12{y_1} - 32 = 0\\ \Rightarrow \qquad & \frac{{128h}}{{{h^2} + {k^2}}} + \frac{{192k}}{{{h^2} + {k^2}}} = 32\\\Rightarrow \qquad &{h^2} + {k^2} - 4h - 6k = 0\end{align}\]

Using \((x,y)\) instead of \((h,k)\) we obtain the required locus as

\[{x^2} + {y^2} - 4x - 6y = 0\]

TRY YOURSELF - III

Q. 1 Find the equation of the circle which intersects \({x^2} + {y^2} - 6x + 4y - 3 = 0\)  orthogonally, passes through \((3, 0)\) and touches the \(y\)-axis.

Q. 2 At what angle do the circles

\[\begin{align}{S_1}:{x^2} + {y^2} - 4x + 6y + 11 = 0\\{S_2}:{x^2} + {y^2} - 2x + 8y + 13 = 0\end{align}\]

intersect?

Q. 3 Find the co-ordinates of the point from which the tangents drawn to the following three circles are of equal lengths:

\[\begin{align} & {S_1}:3{x^2} + 3{y^2} + 4x - 6y - 1 = 0\\  & {S_2}:2{x^2} + 2{y^2} - 3x - 2y - 4 = 0\\ &{S_3}:2{x^2} + 2{y^2} - x + y - 1 = 0\end{align}\]

Q. 4 Find the locus of the centre of the circle passing through \((a,b)\) and orthogonal to \({x^2} + {y^2} = {k^2}.\)

Q. 5 Let \(A({a_1},{b_1})\,\) and \(B({a_2},{b_2})\) be two fixed points and \(O\) be the origin. Circles are drawn on OA and OB as diameters. Prove that the length of the common chord is \(\begin{align}\frac{{\left| {{a_1}{b_2} - {a_2}{b_1}} \right|}}{{AB}}\end{align}\) .