Question

For the circles $ {{x}^{2}}+{{y}^{2}}+2\lambda x+c=0,{{x}^{2}}+{{y}^{2}}+2\mu y-c=0 $ , the number of common tangents is; \[\]
A. one\[\]
B. two\[\]
C. three\[\]
D. four\[\]

Answer 1

Hint: We find the centre $ \left( -g,-f \right) $ and radius $ r=\sqrt{{{g}^{2}}+{{f}^{2}}-c} $ by comparing the given equation of circle general equation of circle $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ . We recall that two circles cut each other orthogonally when the distance between centres $ d $ and the radii are related as $ {{d}^{2}}=r_{1}^{2}+r_{2}^{2} $ . Two intersecting circles will always have two common tangents.\[\]

Complete step by step answer:
We know that the general equation of circle in two variables from is given by
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
The coordinates of centre of general circle is given by $ \left( -g,-f \right) $ and the length of the radius of circle as $ r=\sqrt{{{g}^{2}}+{{f}^{2}}-c} $ .
We know that when the square of distance between two centres of two circles is the sum of squares of the radii that is, then circles intersect orthogonally. We are given the following equation of circles
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}+2\lambda x+c=0.........\left( 1 \right) \\
 & {{x}^{2}}+{{y}^{2}}+2\mu y-c=0...........\left( 2 \right) \\
\end{align}\]
Let us denote the centres of circle (1) and (2) as $ {{C}_{1}},{{C}_{2}} $ and the radii of the circle (1) and (2) as $ {{r}_{1}},{{r}_{2}} $ . We compare the equation of circle (1) with the general equation of circle and find $ g=\lambda ,f=0 $ . The coordinate of the centre and radius circle (1) is respectively
\[\begin{align}
  & {{C}_{1}}\left( -g,-f \right)={{C}_{1}}\left( -\lambda ,0 \right) \\
 & {{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{\left( -\lambda \right)}^{2}}+{{0}^{2}}-c}=\sqrt{{{\lambda }^{2}}-c} \\
\end{align}\]

We compare the equation of circle (2) with the general equation of circle and find $ g=0,f=\mu ,c=-c $ . The coordinate of the centre and radius circle (2) is respectively
\[\begin{align}
  & {{C}_{2}}\left( -g,-f \right)={{C}_{1}}\left( 0,-\mu \right) \\
 & {{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{0}^{2}}+{{\left( \mu \right)}^{2}}-\left( -c \right)}=\sqrt{{{\mu }^{2}}+c} \\
\end{align}\]
Let us find the distance between two centres using distance formula,
\[{{C}_{1}}{{C}_{2}}=\sqrt{{{\left( 0-\left( -\lambda \right) \right)}^{2}}+{{\left( -\mu -0 \right)}^{2}}}=\sqrt{{{\lambda }^{2}}+{{\mu }^{2}}}\]
We square $ {{r}_{1}},{{r}_{2}} $ and add them to have;
\[r_{1}^{2}+r_{2}^{2}={{\left( \sqrt{{{\lambda }^{2}}-c} \right)}^{2}}+{{\left( \sqrt{{{\mu }^{2}}+c} \right)}^{2}}={{\lambda }^{2}}-c+{{\mu }^{2}}+c={{\lambda }^{2}}+{{\mu }^{2}}={{\left( {{C}_{1}}{{C}_{2}} \right)}^{2}}\]
Since we have $ {{\left( {{C}_{1}}{{C}_{2}} \right)}^{2}}=r_{1}^{2}+r_{2}^{2} $ the circles intersect orthogonally. We know that intersecting circles have 2 common tangents. Hence the correct option is B. \[\]

Note:
 We note that intersecting orthogonally means intersecting at right angles. We can alternatively solve using the fact that two circles have 4 common tangents when $ {{C}_{1}}{{C}_{2}} > {{r}_{1}}+{{r}_{2}} $ without touching each other externally , 3 common tangents when $ {{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}} $ touching each other externally, 2 common tangents when $ \left| {{r}_{1}}-{{r}_{2}} \right| < {{C}_{1}}{{C}_{2}} < {{r}_{1}}+{{r}_{2}} $ intersecting each other ,1 common tangent when $ {{C}_{1}}{{C}_{2}} < \left| {{r}_{1}}-{{r}_{2}} \right| $ touching each other internally and no common tangent if $ {{C}_{1}}{{C}_{2}} < {{r}_{1}}+{{r}_{2}} $ not touching each other internally.