# Find the Points Where the Line Cuts the Circle

To find the points where the given line cuts the circle, we take an example as follows:

Example: Find the points where the line $3x + y – 5 = 0$ cuts the given circle ${x^2} + {y^2} = 25$. Also the length of the chord cut off form line by circle.

We have the given line and the circle

$\begin{gathered} 3x + y – 5 = 0\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ {x^2} + {y^2} = 25\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

First we find the points of intersection of the line (i) and the given circle (ii) by using the method of solving simultaneous equations, and from the line (i) we take the variable $y$ separate as follows:

$y = 5 – 3x\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

By putting the value of $y$ from (iii) in equation (ii) we get these results:

$\begin{gathered} {x^2} + {\left( {5 – 3x} \right)^2} = 25 \\ \Rightarrow {x^2} + 25 – 30x + 9{x^2} = 25 \\ \Rightarrow 10{x^2} – 30x = 0 \\ \Rightarrow 10x\left( {x – 3} \right) = 0 \\ \Rightarrow x = 0,\,\,\,x – 3 = 0 \\ \Rightarrow x = 0,\,\,\,x = 3 \\ \end{gathered}$

By putting $x = 0$ in equation (iii), we have $y = 5$. This shows that one point of the intersection of the line and circle is $A\left( {0,5} \right)$. Next, by putting $x = 3$ in equation (iii) again we have $y = – 4$. This shows that the other point of intersection of the line and circle is $B\left( {3, – 4} \right)$.

The length of the chord cut off from the line and the circle is given as using distance formula applying on these two points
$\begin{gathered} \left| {AB} \right| = \sqrt {{{\left( {0 – 3} \right)}^2} + {{\left( {5 – \left( { – 4} \right)} \right)}^2}} \\ \,\,\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( { – 3} \right)}^2} + {{\left( 9 \right)}^2}} = \sqrt {9 + 81} \\ \,\,\,\,\,\,\,\,\,\,\, = \sqrt {90} = \sqrt {9 \times 10} = 3\sqrt {10} \\ \end{gathered}$