A line and a circle

A line and a circle meet at two points, at one point or at none:

• the line cuts the circle at two points – $\displaystyle {y}={1}$ in the drawing;
• the line is a tangent to the circle – it touches it at one point, the point of contact – $\displaystyle {y}={3}$ in the drawing;
• the line does not meet the circle – they have no common point – $\displaystyle {y}={4}$ in the drawing.

The position is found in two ways: by the centre’s distance from the line, or by the number of solutions of a system of equations.

Compare the radius $\displaystyle R$ with the distance $\displaystyle d$ of the circle’s centre from the line:

• $\displaystyle d<R$ – the line cuts the circle;
• $\displaystyle {d}={R}$ – the line is a tangent to the circle;
• $\displaystyle d>R$ – the line does not meet the circle.

The distance of a point from a line is the length of the perpendicular from it to the line. It is found in three steps: the equation of the perpendicular from the centre to the line, the point where the perpendicular meets the line, and the distance between the centre and that point.

Example: the centre of the circle is $\displaystyle K(3,1)$, the radius $\displaystyle 5$, and the line $\displaystyle {y}={2x}$.

• The line’s slope is $\displaystyle 2$, so the perpendicular’s slope is $\displaystyle -\frac{1}{2}$. The perpendicular passes through $\displaystyle K$: $\displaystyle {y-1}={-\frac{1}{2}(x-3)}$, that is $\displaystyle {y}={-\frac{1}{2}x+2.5}$.
• Where they meet: $\displaystyle {2x}={-\frac{1}{2}x+2.5}$, so $\displaystyle {x}={1}$, $\displaystyle {y}={2}$. This is the point $\displaystyle H(1,2)$.
• The distance: $\displaystyle {KH}={\sqrt{(3-1)^{2}+(1-2)^{2}}}={\sqrt{5}\approx2.24}$.

$\displaystyle \sqrt{5}<5$, so the line cuts the circle.

A point common to the line and the circle satisfies both equations. So we solve the system of equations, and the number of solutions is the number of meeting points: two – it cuts, one – a tangent, none – it does not meet the circle.

The same example: $\displaystyle {(x-3)^{2}+(y-1)^{2}}={25}$ and $\displaystyle {y}={2x}$. We substitute $\displaystyle {y}={2x}$ in the circle’s equation:

The solutions: $\displaystyle {x}={3}$ or $\displaystyle {x}={-1}$, and from $\displaystyle {y}={2x}$ the points $\displaystyle (3,6)$ and $\displaystyle (-1,-2)$. Two points – the line cuts the circle, as the first way found.

The centre of the circle is $\displaystyle K(3,-2)$, and the radius $\displaystyle \sqrt{20}$.
The line: $\displaystyle {y}={-2x+6}$.
Find the position of the circle and the line by the centre’s distance from the line.

Write the equation of the perpendicular from $\displaystyle K$ to the line, in a row of your own on the board.

What is the position of the circle and the line? Explain.

The centre of the circle is $\displaystyle (3,-1)$, and the radius $\displaystyle 4$.
The line: $\displaystyle {y}={-x+10}$.

What is the position of the circle and the line? Explain by the number of solutions.

The circle: $\displaystyle {x^{2}+y^{2}}={10}$.
The line: $\displaystyle {y}={3x-10}$.

What is the position of the circle and the line? If they have common points, write them.

The line $\displaystyle {y}={7}$ cuts the circle $\displaystyle {(x+1)^{2}+(y-3)^{2}}={25}$ at the points $\displaystyle A$ and $\displaystyle B$.
Find the coordinates of $\displaystyle A$ and $\displaystyle B$.

The centre of the circle is $\displaystyle (2,-1)$, and the radius $\displaystyle 3$.
The line: $\displaystyle {y}={2}$.

What is the position of the circle and the line? Explain.