Question

If the two circles ${(x-1)}^2+{(y-3)}^2=r^2$ and $x^2+y^2-8x+2y+8=0$ intersect in two distinct points, then

• A. $2<r<8$
• B. $r<2$
• C. $r=2$
• D. $r>2$

Answer 1

The correct option is A

$2<r<8$

Explanation for the correct option.

Step 1. Find the center and radius of the two circle.

On comparing the equation of circle ${(x-1)}^2+{(y-3)}^2=r^2$ with its standard form it can be said that its center is $C_1(1,3)$ and radius is $r_1=r$.

Writing the equation $x^2+y^2-8x+2y+8=0$ in standard form:

$\begin{array}{rcl}(x^2-8x)+(y^2+2y)+8& =& 0\\ \Rightarrow (x^2-2·x·4+4^2)+(y^2+2·y·1+1^2)+8-4^2-1^2& =& 0\\ & \Rightarrow & {(x-4)}^2+{(y+1)}^2=-8+16+1\\ {\Rightarrow (x-4)}^2+{(y+1)}^2& =& 9\\ & \Rightarrow & {(x-4)}^2+{(y+1)}^2=3^2\end{array}$

So comparing it with standard form the center of the second circle is $C_2(4,-1)$ and its radius is $r_2=3$.

Step 2. Find the distance between the two centers.

The distance between $C_1(1,3)$ and $C_2(4,-1)$ is given as:

$$\begin{gathered} C_1{C}_2=\sqrt{{(4-1)}^2+{(-1-3)}^2} \\ \begin{array}{rcl}& \Rightarrow & \end{array}{C}_1{C}_2=\sqrt{{\left(3\right)}^2+{(-4)}^2} \\ \begin{array}{rcl}& \Rightarrow & \end{array}{C}_1{C}_2=\sqrt{9+16} \\ \begin{array}{rcl}& \Rightarrow & \end{array}{C}_1{C}_2=\sqrt{25} \\ \begin{array}{rcl}& \Rightarrow & \end{array}{C}_1{C}_2=5 \end{gathered}$$

Step 3. Forming the inequality and solving it.

As the two circles intersect at two distinct points so the distance between the two centers lies between $\left|r_1-r_2\right|$ and $\left|r_1+r_2\right|$.

$\begin{array}{rcl}\left|r_1-r_2\right|& <& C_1{C}_2<\left|r_1+r_2\right|\\ r-3& <& 5<r+3\end{array}$

Splitting the compound inequalities we get:

$\begin{array}{rcl}r-3& <& 5\\ r-3+3& <& 5+3\\ r& <& 8\end{array}$

and

$\begin{array}{rcl}r+3& >& 5\\ r+3-3& >& 5-3\\ r& >& 2\end{array}$

Therefore, the condition for $r$ is $2<r<8$.

Hence, the correct option is A.