Question

The circles \[{{x}^{2}}+{{y}^{2}}+2x+4y-20=0\] and \[{{x}^{2}}+{{y}^{2}}+6x-8y+10=0\]
(a) are such that the number of common tangents on them is 2.
(b) are orthogonal
(c) are such that the length of their common tangent is \[5{{\left( \dfrac{12}{5} \right)}^{\dfrac{1}{4}}}\].
(d) are such that the length of their common chord is \[5\sqrt{\dfrac{3}{2}}\].

Answer 1

Hint: Assume the center and radius of first circle as \[{{C}_{1}}\] and \[{{r}_{1}}\] and that of second circle as \[{{C}_{2}}\] and \[{{r}_{2}}\] respectively. For option (a), find the value of \[{{r}_{1}}\] and \[{{r}_{2}}\] and that of \[{{C}_{1}}{{C}_{2}}\]. \[{{C}_{1}}{{C}_{2}}\] is the distance between centers. If \[{{r}_{1}}+{{r}_{2}}>{{C}_{1}}{{C}_{2}}\] then option (a) will be correct.

Complete step-by-step solution
For option (b), check the condition:- \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\]. Here, \[\left( -{{g}_{1}},-{{f}_{1}} \right)\] are x and y – coordinate of center of first circle while \[\left( -{{g}_{2}},-{{f}_{2}} \right)\] are x and y – coordinate of center of second circle. \[{{c}_{1}}\] and \[{{c}_{2}}\]are the constant terms in the equations of first and second circle respectively.
For option (c), find the length of common tangent by using the formula, \[l=\sqrt{{{\left( {{C}_{1}}{{C}_{2}} \right)}^{2}}+{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}\], where, l = length of tangent.
For option (d), find the length of common chord by using the formula, \[x=\dfrac{2{{r}_{1}}\times {{r}_{2}}}{{{C}_{1}}{{C}_{2}}}\], where x = length of common chord.
We have been provided with two equations : - \[{{x}^{2}}+{{y}^{2}}+2x+4y-20=0\] and \[{{x}^{2}}+{{y}^{2}}+6x-8y+10=0\]. Let us assume that \[{{x}^{2}}+{{y}^{2}}+2x+4y-20=0\] is the equation of first circle with center \[{{C}_{1}}\] and radius \[{{r}_{1}}\]. So, \[{{x}^{2}}+{{y}^{2}}+6x-8y+10=0\] is the equation of second circle with center \[{{C}_{2}}\] and radius \[{{r}_{2}}\].
To find the coordinates of the center of the circles we compare it with the general equation of the circle, \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\], where center is (-g, -f). Therefore, on comparing above two equations with the general form, we get, \[{{C}_{1}}=\left( -{{g}_{1}},-{{f}_{1}} \right)=\left( -1,-2 \right)\] and \[{{C}_{2}}=\left( -{{g}_{2}},-{{f}_{2}} \right)=\left( -3,4 \right)\].
Now, radius = \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].
\[\begin{align}
  & \Rightarrow {{r}_{1}}=\sqrt{{{g}_{1}}^{2}+{{f}_{1}}^{2}-{{c}_{1}}}=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}-\left( -20 \right)}=\sqrt{25}=5 \\
 & \Rightarrow {{r}_{2}}=\sqrt{{{g}_{2}}^{2}+{{f}_{2}}^{2}-{{c}_{2}}}=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 4 \right)}^{2}}-10}=\sqrt{15} \\
\end{align}\]
(a) Now, \[{{r}_{1}}+{{r}_{2}}=5+\sqrt{15}\], and \[{{C}_{1}}{{C}_{2}}=\] distance between the centers of the circle. Therefore, using distance formula, we get,
\[\begin{align}
  & \Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{{{\left[ -3-\left( -1 \right) \right]}^{2}}+{{\left[ 4-\left( -2 \right) \right]}^{2}}} \\
 & \Rightarrow {{C}_{1}}{{C}_{2}}=\sqrt{36+4}=\sqrt{40} \\
\end{align}\]
Clearly, \[{{r}_{1}}+{{r}_{2}}>{{C}_{1}}{{C}_{2}}\]. Therefore, the two circles will intersect each other at two distinct points. Hence, there will be two common tangents.
So, option (a) is correct.
(b) Orthogonal circles means, at point of intersection the tangents drawn to the two circles are perpendicular to each other. Condition for orthogonality is: - \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\]
Now, L.H.S = \[2\times \left( -3 \right)\times \left( -1 \right)+2\times \left( -2 \right)\times \left( 4 \right)\]
L.H.S = $+6 + (-16)$
L.H.S = $6 – 16 $
L.H.S = $-10$
Also, R. H. S = \[{{c}_{1}}+{{c}_{2}}\]
R. H. S = $-20 + 10$
R. H. S = $-10$
Clearly, \[2{{g}_{1}}{{g}_{2}}+2{{f}_{1}}{{f}_{2}}={{c}_{1}}+{{c}_{2}}\], therefore the circles are orthogonal.
So, option (b) is correct.
(c) Here, we have to find the length of common tangent which is given by: - \[l=\sqrt{{{\left( {{C}_{1}}{{C}_{2}} \right)}^{2}}+{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}}\], where l = length of common tangent.
\[\begin{align}
  & \Rightarrow l=\sqrt{{{\left( \sqrt{40} \right)}^{2}}-{{\left( 5-\sqrt{15} \right)}^{2}}} \\
 & \Rightarrow l=\sqrt{40-\left( 25+15-10\sqrt{15} \right)} \\
 & \Rightarrow l=\sqrt{10\sqrt{15}} \\
 & \Rightarrow l={{10}^{\dfrac{1}{2}}}\times {{\left( \sqrt{15} \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow l={{2}^{\dfrac{1}{2}}}\times {{5}^{\dfrac{1}{2}}}\times {{3}^{\dfrac{1}{4}}}\times {{5}^{\dfrac{1}{4}}} \\
\end{align}\]
This can be simplified as: -
\[\begin{align}
  & \Rightarrow l=\left( {{4}^{\dfrac{1}{4}}}\times {{3}^{\dfrac{1}{4}}} \right)\times \dfrac{{{5}^{\dfrac{1}{2}}}\times {{5}^{\dfrac{1}{4}}}\times {{5}^{\dfrac{1}{4}}}}{{{5}^{\dfrac{1}{4}}}} \\
 & \Rightarrow l=5{{\left( \dfrac{12}{5} \right)}^{\dfrac{1}{4}}} \\
\end{align}\]
So, option (c) is correct.
(d) At last, we have to find the length of common chord which is given by the formula: - \[x=\dfrac{2{{r}_{1}}\times {{r}_{2}}}{{{C}_{1}}{{C}_{2}}}\], where x = length of common chord.
\[\begin{align}
  & \Rightarrow x=\dfrac{2\times 5\times \sqrt{15}}{\sqrt{40}} \\
 & \Rightarrow x=\dfrac{10\times \sqrt{3}\times \sqrt{5}}{\sqrt{8}\times \sqrt{5}} \\
 & \Rightarrow x=\dfrac{10\sqrt{3}}{2\sqrt{2}} \\
 & \Rightarrow x=5\sqrt{\dfrac{3}{2}} \\
\end{align}\]
So, option (d) is correct. Hence, all the options are correct.

Note: One may note that the above formulas we are applying can be derived easily by considering two circles. But here we have not derived them and applied directly because, in this type of question where all four options can be correct, we do not have time to derive such formulas. So, one should remember these basic theories and formulas to solve the questions in less time.