Question

The number of common tangents of the circles given by $$x^2+y^2-8x-2y+1=0$$ and $$x^2+y^2+6x+8y=0$$ is
A. one
B. three
C. two
D. four

Answer 1

Hint: The general equation of the circle $$x^2+y^2+2ax+2by+c=0$$, where $$\left(-a,-b\right)$$ is the center of the circle. Apply this formula, and then use the given conditions to find the required values. Then we will find the distance between the two points $$\left(x_1,y_1\right)$$ and $$\left(x_2,y_2\right)$$ is $$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$$. Then the sum of the radius and distance between the centers of circles are compared to find the number of common tangents.

Complete step by step answer:
Given two circles are
$$x^2+y^2-8x-2y+1=0\text{ ......}\left(1\right)$$
$$x^2+y^2+6x+8y=0\text{ ......}\left(2\right)$$

We know that the general equation of the circle $$x^2+y^2+2ax+2by+c=0$$, where $$\left(-a,-b\right)$$ is the center of the circle.

First, we will take the equation $$\left(1\right)$$.

Comparing the equation $$\left(1\right)$$ and the general equation of the circle to find the center of the circle, we get

$$\left(-a,-b\right)=\left(4,1\right)$$

We know that the formula to find the radius of a circle is $$\sqrt{a^2+b^2+c}$$.

Substituting the value of $$a$$, $$b$$ and $$c$$ in the above formula of radius, we get

$$\begin{gathered} \text{Radius}=\sqrt{4^2+1^2-1} \\ =\sqrt{16+1-1} \\ =\sqrt{16} \\ =4 \end{gathered}$$

Now we will take the equation $$\left(2\right)$$.

Comparing the equation $$\left(2\right)$$ and the general equation of the circle to find the center of the circle, we get

$$\left(-a,-b\right)=\left(-3,-4\right)$$

We know that the formula to find the radius of a circle is $$\sqrt{a^2+b^2+c}$$.

Substituting the value of $$a$$, $$b$$ and $$c$$ in the above formula of radius, we get

$$\begin{gathered} \text{Radius}=\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2+0} \\ =\sqrt{9+16} \\ =\sqrt{25} \\ =5 \end{gathered}$$

Since we know that the distance between the two points $$\left(x_1,y_1\right)$$ and $$\left(x_2,y_2\right)$$ is $$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$$.

Now we will find the distance between the center of the two circles $$\left(4,1\right)$$ and $$\left(-3,-4\right)$$, where $$x_1=4$$, $$y_1=1$$, $$x_2=-3$$ and $$y_2=-4$$.

$$\begin{gathered} \sqrt{{\left(4-\left(-3\right)\right)}^2+{\left(1-\left(-4\right)\right)}^2}=\sqrt{7^2+5^2} \\ =\sqrt{49+25} \\ =\sqrt{74} \end{gathered}$$

Now we will find the sum of the radius of two circles.

$$5+4=9$$

As we can see that the distance between the centers is less than the sum of the radius of two circles.
Hence, the circles are intersecting with each other at two points, so the number of common tangents is 2.

Therefore, the option C is correct.

Note: In solving these types of questions, we have to compare the given equations with general equation of the circle $$x^2+y^2+2ax+2by+c=0$$, where $$\left(-a,-b\right)$$ is the center of the circle to find the centers of the circles easily and radius of a circle is $$\sqrt{a^2+b^2+c}$$. After that the question will become really simple to solve.