Circles
12.0 Equation of common tangents
12.0 Equation of common tangents
Let the two circles be $${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r_1}^2...(1)$$ and $${\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} = {r_2}^2...(2)$$
with centres ${C_1}({x_1},{y_1})$ and ${C_2}({x_2},{y_2})$ and radii ${r_1}$ and ${r_2}$ respectively.
CASE I: When $\left| {{C_1}{C_2}} \right| > {r_1} + {r_2}$ i.e., when two circles do not intersect each other.
In this case all four tangents are real and distinct.
Coordinates of point $D$ (intersection of direct common tangents) $$ \equiv \left( {\frac{{{r_1}{x_2} - {r_2}{x_1}}}{{{r_1} - {r_2}}},\frac{{{r_1}{y_2} - {r_2}{y_1}}}{{{r_1} - {r_2}}}} \right)$$
Because $D$ divides ${C_1}$ and ${C_2}$ in the ratio ${r_1}:{r_2}$ externally.
Coordinates of point $T$ (intersection of transverse common tangents) $$ \equiv \left( {\frac{{{r_1}{x_2} + {r_2}{x_1}}}{{{r_1} + {r_2}}},\frac{{{r_1}{y_2} + {r_2}{y_1}}}{{{r_1} + {r_2}}}} \right)$$
Because $T$ divides ${C_1}$ and ${C_2}$ in the ratio ${r_1}:{r_2}$ internally.
(a) To find the equation of direct common tangents:
Let equation of common tangent through $D(\alpha ,\beta )$ is $$y - \beta = m\left( {x - \alpha } \right)...(3)$$
Now, perpendicular from ${C_1}$ or ${C_2}$ on equation $(3)$ equals to ${r_1}$ or ${r_2}$. Then we get the two values of $m$, substitute the values of $m$ in $(1)$, we get two direct common tangents.
(b) To find the equation of transverse common tangents:
Let equation of common tangent through $T\left( {\Upsilon ,\delta } \right)$ is $$y - \delta = m\left( {x - \Upsilon } \right)...(4)$$
Now, perpendicular from ${C_1}$ or ${C_2}$ on equation $(4)$ equals to ${r_1}$ or ${r_2}$. Then we get the two values of $m$, substitute the values of $m$ in $(1)$, we get two transverse common tangents.
CASE II: When $\left| {{C_1}{C_2}} \right| = {r_1} + {r_2}$ i.e., when two circles touch each other externally.
In this case, two direct common tangents are real and distinct while transverse tangents are coincident.
Coordinates of point $D$ (intersection of direct common tangents) $$ \equiv \left( {\frac{{{r_1}{x_2} - {r_2}{x_1}}}{{{r_1} - {r_2}}},\frac{{{r_1}{y_2} - {r_2}{y_1}}}{{{r_1} - {r_2}}}} \right)$$
Because $D$ divides ${C_1}$ and ${C_2}$ in the ratio ${r_1}:{r_2}$ externally.
Coordinates of point $T$ (intersection of transverse common tangents) $$ \equiv \left( {\frac{{{r_1}{x_2} + {r_2}{x_1}}}{{{r_1} + {r_2}}},\frac{{{r_1}{y_2} + {r_2}{y_1}}}{{{r_1} + {r_2}}}} \right)$$
Because $T$ divides ${C_1}$ and ${C_2}$ in the ratio ${r_1}:{r_2}$ internally.
(a) To find the equation of direct common tangents:
Let equation of common tangent through $D(\alpha ,\beta )$ is $$y - \beta = m\left( {x - \alpha } \right)...(3)$$
Now, perpendicular from ${C_1}$ or ${C_2}$ on equation $(3)$ equals to ${r_1}$ or ${r_2}$. Then we get the two values of $m$, substitute the values of $m$ in $(1)$, we get two direct common tangents.
(b) To find the equation of transverse common tangents:
Equation of circles are $${S_1} \equiv {\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} - {r_1}^2 = 0$$ and $${S_2} \equiv {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} - {r_2}^2 = 0$$
then, equation of common tangent is ${S_1} - {S_2} = 0$ which is same as equation of common chord.
CASE III: When $\left| {{r_1} - {r_2}} \right| < \left| {{C_1}{C_2}} \right| < {r_1} + {r_2}$ i.e., when two circles intersect each other.
In this case two direct common tangents are real and distinct while the transverse tangents are imaginary.
(a) To find the equation of direct common tangents:
Let equation of common tangent through $D(\alpha ,\beta )$ is $$y - \beta = m\left( {x - \alpha } \right)...(3)$$
Now, perpendicular from ${C_1}$ or ${C_2}$ on equation $(3)$ equals to ${r_1}$ or ${r_2}$. Then we get the two values of $m$, substitute the values of $m$ in $(1)$, we get two direct common tangents.
CASE IV: When $\left| {{C_1}{C_2}} \right| = \left| {{r_1} - {r_2}} \right|$ i.e., when two circles touch each other internally.
In this case two tangents are coincident while the other two tangents are imaginary.
Equation of circles are $${S_1} \equiv {\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} - {r_1}^2 = 0$$ and $${S_2} \equiv {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} - {r_2}^2 = 0$$
then, equation of common tangent is ${S_1} - {S_2} = 0$ which is same as equation of common chord.
CASE V: When $\left| {{C_1}{C_2}} \right| < \left| {{r_1} - {r_2}} \right|$ i.e., when two circles neither cut nor touch each other.
In this case all the four common tangents are imaginary.
Question 22. Find the number of common tangents of the two circles ${S_1} \equiv {x^2} + {y^2} - 2x - 6y + 9 = 0$ and ${S_2} \equiv {x^2} + {y^2} + 6x - 2y + 1 = 0$ and also find the equation of tangents.
Solution: Centre of circle ${S_1}$ is ${C_1}(1,3)$ and radius ${r_1} = 1$. Centre of circle ${S_2}$ is ${C_2}(-3,1)$ and radius ${r_2} = 3$.
Distance between centre ${C_1}$ and ${C_2}$ i.e., ${C_1}{C_2} = \sqrt {{{\left( {1 + 3} \right)}^2} + {{\left( {3 - 1} \right)}^2}} = 2\sqrt 5 $.
Sum of radius ${r_1} + {r_2} = 1 + 3 = 4$. As $\left| {{C_1}{C_2}} \right| > {r_1} + {r_2}$, two circles are non-intersecting. Therefore, number of common tangents is $4$.
Let us assume the intersection point of two direct common tangents be $D(\alpha ,\beta )$ and the intersection point of two transverse common tangents be $T(\Upsilon ,\delta )$.
To find the equation of transverse common tangent:
As shown in figure $42$, $\Delta TP{C_2} \cong \Delta TQ{C_1}$
Therefore, $$\frac{{{C_1}T}}{{{C_2}T}} = \frac{{{r_1}}}{{{r_2}}}$$
Use section formulae between points ${C_1}$ and ${C_2}$, we get $$\Upsilon = \frac{{ - 3 + 3}}{4} = 0$$ and $$\delta = \frac{{1 + 9}}{4} = \frac{5}{2}$$
The coordinates of point $T$ is $(0,\frac{5}{2})$.
The equation of line having slope $m$ and passing through $T$ can be written as $$y - \frac{5}{2} = m(x - 0)$$ $$y - mx - \frac{5}{2} = 0$$
Now, the perpendicular distance from centre ${C_1}(1,3)$ to the line $(1)$ is equal to radius of circle ${r_1}$ $$1 = \left| {\frac{{3 - m - \frac{5}{2}}}{{\sqrt {1 + {m^2}} }}} \right|$$
Squaring both sides we get, $$1 + {m^2} = {(\frac{1}{2} - m)^2}$$ $$1 + {m^2} = \frac{1}{4} + {m^2} - m$$ $$m = \frac{1}{4} - 1 = - \frac{3}{4}{\text{ and }}m = \infty $$
Therefore, equations of transverse common tangents are $$y + \frac{3}{4}x = \frac{5}{2}$$ and $$y = \frac{5}{2}$$
To find the equation of direct common tangent:
As shown in figure $43$, $\Delta DR{C_2} \cong \Delta DS{C_1}$
Therefore, $$\frac{{{C_1}D}}{{{C_2}D}} = \frac{{{r_1}}}{{{r_2}}}$$
Use section formulae between points ${C_1}$ and ${C_2}$, we get $$\alpha = \frac{{3 + 3}}{2} = 3$$ and $$\beta = \frac{{9 - 1}}{2} = 4$$
The coordinates of point $D$ is $(3,4)$.
The equation of line having slope $m'$ and passing through $D$ can be written as $$y - 4 = m'(x - 3)$$ or, $$y - m'x + 3m' - 4 = 0...(1)$$
Now, the perpendicular distance from centre ${C_1}(1,3)$ to the line $(1)$ is equal to radius of circle ${r_1}$ $$1 = \left| {\frac{{3 - m' + 3m' - 4}}{{\sqrt {1 + {{m'}^2}} }}} \right|$$
Squaring both sides we get, $$1 + {m'^2} = {(2m' - 1)^2}$$ $$1 + {m'^2} = 4{m'^2} + 1 - 4m'$$ $$3{m'^2} - 4m' = 0$$ $$m'\left( {3m' - 4} \right) = 0$$
$$m' = 0{\text{ and }}m' = \frac{4}{3}$$
Therefore, equations of direct common tangents are $$y = \frac{4}{3}x{\text{ and }}y = 4$$
