Line and circle contact
Substitution changes a line and circle problem into a quadratic equation. The discriminant counts the common points, while a tangent radius is perpendicular to the tangent line
- Substitute the line Replace the line variable in the circle equation to obtain a quadratic \(\displaystyle \left(x-1\right)^2+\left(y+2\right)^2=25,\ y=2x+1\Rightarrow\left(x-1\right)^2+\left(2x+3\right)^2=25\)
- Read the discriminant A negative, zero or positive discriminant gives zero, one or two common points respectively \(\displaystyle \Delta<0\Rightarrow0,\quad\Delta=0\Rightarrow1,\quad\Delta>0\Rightarrow2\)
- Use perpendicular gradients When both gradients exist, the radius and tangent gradients are negative reciprocals \(\displaystyle m_r=2\Rightarrow m_t=-\frac{1}{m_r}=-\frac{1}{2}\)
- Subtract squared lengths In the right triangle, the tangent length is found after subtracting the squared radius from the squared distance to the external point \(\displaystyle CP=13,\ CT=5\Rightarrow PT^2=CP^2-CT^2=13^2-5^2=144\)
Worked example
The circle and line are given below. Determine how many common points they have
- \(\displaystyle \left(x-2\right)^2+\left(3x-3\right)^2=25\)Substitute the line equation into the circle equation
- \(\displaystyle 10x^2-22x-12=0\)Expand both squares and collect the terms in x.
- \(\displaystyle \Delta=(-22)^2-4(10)(-12)=964\)Use the quadratic coefficients in the discriminant formula
- \(\displaystyle \Delta=964>0\Rightarrow\text{two distinct common points}\)A positive discriminant gives two distinct real solutions
Common mistakes
Difficulty levels
Questions about this topic
What does the discriminant say about a line and a circle?+
After substitution gives a quadratic, a negative discriminant means no common points, zero means one repeated point, and a positive discriminant means two common points
Why is a tangent perpendicular to the radius?+
The shortest distance from the centre to a line is perpendicular to the line, and the tangent touches the circle at the endpoint of that radius
Why is the tangent length found by subtracting squares?+
The centre, contact point and external point form a right triangle, so subtract the squared radius from the squared centre-to-external-point distance to find the squared tangent length
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 6: Functions and relations