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 \(\begin{align}\frac{{\left| {{a_1}{b_2} - {a_2}{b_1}} \right|}}{{AB}}\end{align}\) .