Intersections of lines and circles

To find the intersection points between a circle and a straight line, substitute the equation of the straight line into the circle's equation and solve for $x$ and $y$. Let’s work through this step by step.

Step 1: Start with the Equations

The equation of a circle is typically given as:

$$x^2+y^2=r^2$$

where $r$ is the radius.

The equation of a straight line is usually written as:

$$y=mx+c$$

where $m$ is the gradient and $c$ is the $y$-intercept.

Step 2: Substitute the Line Equation into the Circle Equation

Replace $y$ in the circle's equation with $mx+c$. For example, consider:

$$x^2+y^2=25\text{and}y=x+3$$

Substitute $y=x+3$ into $x^2+y^2=25$:

$$x^2+(x+3{)}^2=25$$

Step 3: Expand and Simplify

Expand the squared term and simplify the equation:

$$x^2+(x^2+6x+9)=25$$

Combine like terms:

$$2x^2+6x+9=25$$

Simplify further:

$$2x^2+6x-16=0$$

Step 4: Solve the Quadratic Equation

Factorise the quadratic equation:

$$2x^2+6x-16=0$$

Factor out the common factor $2$:

$$2(x^2+3x-8)=0$$

Factorise $x^2+3x-8$:

$$(x+4)(x-2)=0$$

So the solutions are:

$$x=-4\text{and}x=2$$

Step 5: Find the Corresponding $y$-Values

Substitute these $x$-values back into the line equation $y=x+3$:

For $x=-4$:

$$y=-4+3=-1$$

For $x=2$:

$$y=2+3=5$$

The intersection points are:

$$(-4,-1)\text{and}(2,5)$$

Step 6: Verify the Solutions

Check that the points satisfy both the circle and line equations. For example:

For $(-4,-1)$:

$$x^2+y^2=(-4{)}^2+(-1{)}^2=16+1=25$$

For $(2,5)$:

$$x^2+y^2=2^2+5^2=4+25=25$$

Both points satisfy the circle equation.

Summary

• Substitute the straight line equation into the circle equation.
• Expand and simplify to form a quadratic equation.
• Solve the quadratic equation to find $x$-values.
• Substitute $x$-values back into the line equation to find $y$-values.
• Verify the solutions to ensure they satisfy both equations.