Question

The minimum distance between any two points ${\text{P}}_1$ and ${\text{P}}_2$ while considering point ${\text{P}}_1$ on one circle and point ${\text{P}}_2$ one the other circle for the given circles’ equations $x^2+y^2-10x-10y+41=0$ and $x^2+y^2-24x-10y+160=0$ ________

Answer 1

Step 1: Find the radii and centers of the given circles.

The equations of the two circles are given.

$$\begin{gathered} x^2+y^2-10x-10y+41=0...1 \\ {x}^2+y^2-24x-10y+160=0...2 \end{gathered}$$

We know that the general equation of the circle is ${(x-x_1)}^2+{(y-y_1)}^2=r^2...3$.

Where, $(x,y)$ is the general point of the circle.

$(x_1,y_1)$ is the center coordinate.

$r$ is the radius of the circle.

Rewrite the equation $1$ as follows:

$$\begin{gathered} x^2+y^2-10x-10y+25+16+9-9=0 \\ \Rightarrow \left(x^2-10x+25\right)+\left(y^2-10y+25\right)=9 \\ \Rightarrow {\left(x-5\right)}^2+{\left(y-5\right)}^2=3^2...4 \end{gathered}$$

On comparing equation $3$ and equation $4$, we get

The radius of the first circle is $3$ and the coordinates of the center is $(5,5)$.

Rewrite the equation $2$ as follows:

$$\begin{gathered} x^2+y^2-24x-10y+144+16+9-9=0 \\ \Rightarrow \left(x^2-24x+144\right)+\left(y^2-10y+25\right)=9 \\ \Rightarrow {\left(x-12\right)}^2+{\left(y-5\right)}^2=3^2...5 \end{gathered}$$

On comparing equation $3$ and equation $5$, we get,

The radius of the first circle is $3$ and the coordinates of the center is $(12,5)$.

Step 2: Find the minimum distance between the points ${\text{P}}_1$ and ${\text{P}}_2$ .

We know that, the minimum distance between the circles is given by $C_1{C}_2-r_1-r_2$. when the distance between the centers is greater than the sum of radii.

Where, $C_1,C_2$ are the coordinates of the centers of the two circles.

$r_1,r_2$ are the radii of the two circles.

So, the minimum distance between the points ${\text{P}}_1$ and ${\text{P}}_2$ can be given by:

$$\begin{gathered} {\left(P_1{P}_2\right)}_{\text{min}}=\sqrt{{(12-5)}^2+{(5-5)}^2}-3-3 \\ \Rightarrow {\left(P_1{P}_2\right)}_{\text{min}}=\sqrt{7^2+0}-6 \\ \Rightarrow {\left(P_1{P}_2\right)}_{\text{min}}=7-6 \\ \Rightarrow {\left(P_1{P}_2\right)}_{\text{min}}=1 \end{gathered}$$

Therefore, the minimum distance between the points ${\text{P}}_1$ and ${\text{P}}_2$ is $1$ unit.

Hence, the answer is $1$.