Question

Let \[{s_1} = {x^2} + {y^2} = 9\] and \[{s_2} = {\left( {x - 2} \right)^2} + {y^2} = 1\]. What is the locus of the center of a variable circle \[s\] which touches \[{s_1}\] internally and \[{s_2}\] externally always passes through the points
A. \[\left( {\dfrac{1}{2}, \pm \dfrac{{\sqrt 5 }}{2}} \right)\]
B. \[\left( {2, \pm \dfrac{3}{2}} \right)\]
C. \[\left( {1, \pm 2} \right)\]
D. \[\left( {0, \pm \sqrt 3 } \right)\]

Answer 1

Hint: Find the centers and radii of the two given circles. Assume that the center of the variable circle \[s\] be \[\left( {h,k} \right)\] and radius be \[r\] units. Find the distances between the centers of the variable circle and the given circles and add them. The sum will be constant and hence the locus will represent an ellipse having foci at the centers of the given circles. Find the center of the ellipse and it’s equation. After that check which of the given points satisfies the equation.

Formula Used:
Equation of a circle having centre at \[\left( {\alpha ,\beta } \right)\] and radius \[r\] units is \[{\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}\]
Equation of an ellipse having centre at \[\left( {\alpha ,\beta } \right)\], semi-major axis of \[a\] units and semi-minor axis of \[b\] units is \[\dfrac{{{{\left( {x - \alpha } \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - \beta } \right)}^2}}}{{{b^2}}} = 1\], where \[a > b\]
Eccentricity of the ellipse is \[e = \sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} \]
Distance between the foci is \[2ae\] units.
Coordinates of midpoint of the line segment joining the points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)\]

Complete step-by-step answer:
The given circles are \[{s_1} = {x^2} + {y^2} = 9\] and \[{s_2} = {\left( {x - 2} \right)^2} + {y^2} = 1\]
Comparing the equations with the general form of equation of a circle \[{\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}\], we get
Centre of the first circle is \[{C_1}\left( {0,0} \right)\] and radius is \[3\] units.
Centre of the second circle is \[{C_2}\left( {2,0} \right)\] and radius is \[1\] unit.
Let the centre of the variable circle \[s\] be at \[C\left( {h,k} \right)\] and radius be \[r\] units
It is given that the variable circle \[s\] touches the circle \[{s_1}\] internally and the circle \[{s_2}\] externally.
If two circles touch each other internally, then the distance between the centers of the circles is equal to the difference between the radii of the two circles.
So, distance between the centers of the circles \[s\] and \[{s_1}\] is \[C{C_1} = \left( {3 - r} \right)\] units.
If two circles touch each other externally, then the distance between the centers of the circles is equal to the sum of the radii of the two circles.
So, distance between the centers of the circles \[s\] and \[{s_2}\] is \[{C_2}C = \left( {r + 1} \right)\] units.
Now, \[C{C_1} + {C_2}C = \left( {3 - r} \right) + \left( {r + 1} \right) = 3 - r + r + 1 = 4\]
This shows that the sum of the distances of the centers of the two given circles from the centre of the variable circle doesn’t depend on the value of \[r\], it is constant.
It means the locus of the centre of the variable circle \[s\] is an ellipse having major axis of length \[4\] units and foci at \[{C_1}\] and \[{C_2}\].
Centre of the ellipse is the midpoint of \[{C_1}\] and \[{C_2}\].
Midpoint of \[{C_1}\left( {0,0} \right)\] and \[{C_2}\left( {2,0} \right)\] is \[\left( {\dfrac{{0 + 2}}{2},\dfrac{{0 + 0}}{2}} \right) = \left( {1,0} \right)\]
So, centre of the ellipse is at \[\left( {1,0} \right)\].
If \[2a\] be the length of the major axis of the ellipse then
\[2a = 4 \Rightarrow a = 2\]
If \[e\] be the eccentricity of the ellipse then distance between the foci is
\[2ae = {C_1}{C_2} = 3 - 1 = 2\]
\[ \Rightarrow 4e = 2\]
\[ \Rightarrow e = \dfrac{2}{4} = \dfrac{1}{2}\]
If \[b\] be the length of the semi-minor axis, then
\[{b^2} = {a^2}\left( {1 - {e^2}} \right)\]
\[ = {\left( 2 \right)^2}\left\{ {1 - {{\left( {\dfrac{1}{2}} \right)}^2}} \right\}\]
\[ = 4\left( {1 - \dfrac{1}{4}} \right)\]
\[ = 4 \times \dfrac{3}{4}\]
\[ = 3\]
\[\therefore b = \sqrt 3 \]
So, equation of the ellipse is
\[\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{{\left( 2 \right)}^2}}} + \dfrac{{{y^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} = 1\]
\[ \Rightarrow \dfrac{{{{\left( {x - 1} \right)}^2}}}{4} + \dfrac{{{y^2}}}{3} = 1\]
If we put \[x = 2\] and \[y = \pm \dfrac{3}{2}\] in the left-hand side, then
\[\dfrac{{{{\left( {2 - 1} \right)}^2}}}{4} + \dfrac{{{{\left( { \pm \dfrac{3}{2}} \right)}^2}}}{3} = \dfrac{1}{4} + \dfrac{{\dfrac{9}{4}}}{3} = \dfrac{1}{4} + \dfrac{9}{4} \times \dfrac{1}{3} = \dfrac{1}{4} + \dfrac{3}{4} = 1\]
So, clearly, the equation of the ellipse is satisfied by \[x = 2\] and \[y = \pm \dfrac{3}{2}\]
Thus, the variable circle always passes through the point \[\left( {2, \pm \dfrac{3}{2}} \right)\].
Hence option B is correct.

Note: Remember that if two circles touch each other internally, then the distance between the centers of the circles is equal to the difference between the radii of the two circles but if two circles touch each other externally, then the distance between the centers of the circles is equal to the sum of the radii of the two circles. The formula of the distance between two foci and eccentricity of an ellipse are invariant.