Question

Find the equation to the pair of transverse common tangents to the circles ${x^2} + {y^2} - 4x - 10y + 28 = 0$ and ${x^2} + {y^2} + 4x - 6y + 4 = 0$

Answer 1

Hint: First of all we will find the center and the radius by equating the circle equation from the question and when the line is divided into the ratio internally than by using the equation $y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$ here, $m\& n$ is the ratio. Then we will get the equation of the common tangent by using the formula $y - {y_1} = m\left( {x - {x_1}} \right)$ and by solving it we will have the required equation.

Formula used:
General equation of a circle,
${x^2} + {y^2} + 2gx + 2fy + c = 0$
Here, $g,f,c$ are constants.
When the equation of line gets internally divided then we will use the formula, $y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$
Here, $m\& n$ is the ratio.

Complete answer:
So we have the equation given as ${x^2} + {y^2} - 4x - 10y + 28 = 0$ and we will name it ${S_1}$
By equating the equation with the general equation of a circle, we get
$ \Rightarrow 2g = - 4$
And on solving we get
$ \Rightarrow g = - 2$
Similarly, $2f = - 10$
And on solving, we get
$ \Rightarrow f = - 5$
So the center of the circle ${S_1}$ will be $\left( { - g, - f} \right)$,
And will be equal to $\left( {2,5} \right)$ .
Now the radius will be equal to $r = \sqrt {{g^2} + {f^2} - c} $
And on substituting the values we get
\[ \Rightarrow {r_1} = \sqrt {{2^2} + {5^2} - 28} \]
And solving we get
\[ \Rightarrow {r_1} = \sqrt {4 + 25 - 28} \]
So the radius will be \[{r_1} = 1\]
Similarly, we have another equation given as ${x^2} + {y^2} + 4x - 6y + 4 = 0$ and we will name it ${S_2}$
By equating the equation with the general equation of a circle, we get
$ \Rightarrow 2g = 4$
And on solving we get
$ \Rightarrow g = 2$
Similarly, $2f = - 6$
And on solving, we get
$ \Rightarrow f = - 3$
So the center of the circle ${S_2}$ will be $\left( { - g, - f} \right)$,
And will be equal to $\left( { - 2,3} \right)$ .
Now the radius will be equal to $r = \sqrt {{g^2} + {f^2} - c} $
And on substituting the values we get
\[ \Rightarrow {r_2} = \sqrt {{2^2} + {3^2} - 4} \]
And solving we get
\[ \Rightarrow {r_2} = \sqrt {4 + 9 - 4} \]
So the radius will be \[{r_2} = 3\] .
So the center will be $\left( {{c_1},{c_2}} \right)$
Hence, $d = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $
And on solving it we get
$ \Rightarrow d = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 2} \right)}^2}} $
So it will be,
$ \Rightarrow d = 2\sqrt 5 $
Since we can see that the $d > {r_1} + {r_2}$
And also we know that the line divides it into two parts internally, so substituting the values, we get
$ \Rightarrow x = \dfrac{{3\left( 2 \right) - 1\left( { - 2} \right)}}{{3 - 1}}$
And on solving it we get
$ \Rightarrow x = 4$
Similarly, for $y$ it will be
$ \Rightarrow y = \dfrac{{3\left( 5 \right) - 1\left( 3 \right)}}{{3 - 1}}$
And on solving it we get
$ \Rightarrow y = 6$
Now for another equation, we will find the same as above, so we get
$ \Rightarrow a = \dfrac{{3\left( 2 \right) + 1\left( { - 2} \right)}}{{3 + 1}}$
And on solving it we get
$ \Rightarrow a = 1$
Similarly, for $y$ it will be
$ \Rightarrow b = \dfrac{{3\left( 5 \right) + 1\left( 3 \right)}}{{3 + 1}}$
And on solving it we get
$ \Rightarrow b = \dfrac{9}{2}$
So, we have $c = \left( {1,\dfrac{9}{2}} \right)$ .
Now the equation of direction, common tangent at $\left( {4,6} \right)$ . So by using the formula $y - {y_1} = m\left( {x - {x_1}} \right)$ and substituting the values, we get
$ \Rightarrow y - 6 = m\left( {x - 4} \right)$
And on solving it we get
$ \Rightarrow y - 6 = mx - 4m$
Taking all the terms on one side we get
$ \Rightarrow mx - y - 4m + 6 = 0$
So from the condition of tangency $\left( {2,5} \right)$ , we have
$ \Rightarrow \dfrac{{2m - 5 - 4m + 6}}{{\sqrt {{m^2} + 1} }} = \pm 1$
By doing the cross multiplication and solving it, we get
$ \Rightarrow - 2m + 1 = \sqrt {{m^2} + 1} $
Taking the square on both sides, we get
$ \Rightarrow {\left( { - 2m + 1} \right)^2} = {m^2} + 1$
So on expanding the equation, we get
$ \Rightarrow 4{m^2} - 4m + 1 = {m^2} + 1$
And on solving and taking the common, we have
$ \Rightarrow m\left( {3m - 4} \right) = 0$
And on solving we have $m = 0$ and $m = \dfrac{4}{3}$
Now similarly we will find the equation of direction, common tangent
$ \Rightarrow y - 6 = \dfrac{4}{3}\left( {x - 4} \right)$
And solving it we have the equation as
$ \Rightarrow 4x - 3y + 2 = 0$
Now the equation of the transverse common tangent at $\left( {1,\dfrac{9}{2}} \right)$ , we have
$ \Rightarrow y - \dfrac{9}{2} = m\left( {x - 1} \right)$
And on substituting the values and solving it, we get
$ \Rightarrow 2m - 5 - m + \dfrac{9}{2}$
And from the condition of tangency, we have
$ \Rightarrow \dfrac{{ \Rightarrow 2m - 5 - m + \dfrac{9}{2}}}{{\sqrt {{m^2} + 1} }} = \pm 1$
And on solving the above equation we get
$ \Rightarrow {m^2} - m + \dfrac{1}{4} = {m^2} + 1$
On canceling the like terms, we get
$ \Rightarrow - \dfrac{3}{4}$
So substituting it in the formula, we get
$ \Rightarrow \left( {y - \dfrac{9}{2}} \right) = m\left( {x - 1} \right)$
Putting the value of $m$ , we have
$ \Rightarrow \left( {y - \dfrac{9}{2}} \right) = - \dfrac{3}{4}\left( {x - 1} \right)$
And on solving it we get
$ \Rightarrow 3x + 4y - 21 = 0$
And on taking the constant term to the right side, we get
$ \Rightarrow 3x + 4y = 21$

Hence, the equation will be $3x + 4y = 21$

Note: As we can see that this type of question becomes lengthy so while solving the equations or applying the formula we should be careful during that. Also if the equation does not get solved by factorization then we will use the determinant formula which is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .