Question

Show that the circles touch each other externally. Find the point of contact and equation of their common tangent.
$$x^2+y^2-4x+10y+20=0$$
$$x^2+y^2+8x-6y-24=0$$

Answer 1

Hint: In order to determine the point of contact an equation of the tangent can be compared with the general equation.
We use the general equation of the circle $$x^2+y^2+2gx+2fy+c=0$$ to find the centre and radius of the circle by comparing with the given equation to find the distance between the centres of the two circles and verify that it is equal to the sum of radius of the two circles. i.e., $$C_1{C}_2=r_1+r_2$$.
We know that distance between two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ is given by $$AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$$
The section formula $$P\left({\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},{\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}\right)$$ is used to find the coordinate of the point of contact then, we will find the equation of tangent using this formula $S_1-S_2=0$, where $S_1$ and $S_2$ is the given equation of the circle.

Complete step-by-step solution:
The given question is based on the circle. A circle is a locus of points whose distance from a fixed point is always constant. The fixed point is called the centre of the circle and the constant distance is called radius.
The general equation of the circle is given is $$x^2+y^2+2gx+2fy+c=0$$ where its centre is $$(-g,-f)$$ and radius is $$\sqrt{g^2+f^2-c}$$
The given equation of the circle is $$x^2+y^2-4x+10y+20=0$$ and $$x^2+y^2+8x-6y-24=0$$
Comparing the given equation with general equation, we get
$$\begin{gathered} 2g_1=-4 \\ 2f_1=10 \end{gathered}$$ and $$\begin{gathered} 2g_2=8 \\ 2f_2=-6 \end{gathered}$$ and $$\begin{gathered} c_1=20; \\ {c}_2=-24; \end{gathered}$$
On simplifying, we have
$$\begin{gathered} g_1=-2 \\ {f}_1=5 \end{gathered}$$and $$\begin{gathered} g_2=4 \\ {f}_2=-3 \end{gathered}$$ and $$\begin{gathered} c_1=20 \\ {c}_2=-24 \end{gathered}$$ .
Therefore centre of the circle $$C_1$$ is $$(-g_1,-f_1)=(2,-5)$$ and radius $$r_1=\sqrt{{g_1}^2+{f_1}^2-c_1}=\sqrt{{(-2)}^2+5^2-20}=3$$
And centre of the circle $$C_2$$ is $$(-g_2,-f_2)=(-4,3)$$ and radius $$r_2=\sqrt{{g_2}^2+{f_2}^2-c_2}=\sqrt{4^2+{(-3)}^2-(-24)}=7$$
We know that distance between two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ is given by $$AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$$
Now the distance between the centre $$C_1(2,-5)$$ and $$C_2(-4,3)$$ of the circle is given by $$C_1{C}_2=\sqrt{{(2+4)}^2+{(-5-3)}^2}=10$$
hence, we see that
$$C_1{C}_2=r_1+r_2=10$$.
Therefore, the circle touches externally.
Now the point of contact says $$P$$ divide the line joining the circle $$C_1(2,-5)$$ and $$C_2(-4,3)$$ in $$3:7$$ ratio.
We know that the coordinate of the point $$P$$ dividing the line segment $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ in ratio $$m:n$$ is given by
 $$P\left({\displaystyle \frac{m{x}_2+n{x}_1}{m+n}},{\displaystyle \frac{m{y}_2+n{y}_1}{m+n}}\right)$$.
Therefore, the coordinate of $$P$$is given as
$$P\left({\displaystyle \frac{(3)(-4)+(7)(2)}{10}},{\displaystyle \frac{(3)(3)+(7)(-5)}{10}}\right)=P\left({\displaystyle \frac{1}{5}},{\displaystyle \frac{13}{5}}\right)$$
Now we know that the equation of common tangent when two circle touches externally is given by $$S_1-S_2=0$$, where $$S_1$$ and $$S_2$$ is the given equation of the circle.
Hence we have
 $$(x^2+y^2-4x+10y+20)-(x^2+y^2+8x-6y-24)=0$$
On expanding the brackets, we get
$\Rightarrow$$$(x^2+y^2-4x+10y+20-x^2-y^2-8x+6y+24)=0$$
On further simplification, then
$$\begin{gathered} \Rightarrow -4x+10y+20-8x+6y+24)=0 \\ \Rightarrow -12x+16y+44=0 \end{gathered}$$
Therefore, the equation of common tangent is $$-12x+16y+44=0$$.

Note: Section formula is used to find the coordinate of the points when it is divided in ratio.
The tangent is a line segment which touches through only one point of the circle.
Distance between two points $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ is given by $$AB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$$