Coordinate Geometry – Problems involving intersections of lines and circles

Lines & Circles — Intersections (9709 P1)

Learning objective: Find points of intersection of a line with a circle by substitution and solving the resulting quadratic; interpret the discriminant to determine 0/1/2 intersections.

Key facts

• Circle (standard form): \((x-a)^2+(y-b)^2=r^2\). Centre \((a,b)\), radius \(r\).
• Circle (general form): \(x^2+y^2+gx+fy+c=0\). Centre \(\left(-\tfrac{g}{2},-\tfrac{f}{2}\right)\), radius \(r=\sqrt{\left(\tfrac{g}{2}\right)^2+\left(\tfrac{f}{2}\right)^2-c}\).
• Line forms: \(y=mx+c\), \(Ax+By+C=0\), and vertical \(x=k\).
• Graph–equation link: intersection points of the graphs \(\Leftrightarrow\) simultaneous solutions of the equations.

Method 1 (coordinates via substitution)

1. Write the line as \(y=mx+c\) (or handle \(x=k\) separately).
2. Substitute into the circle to get a quadratic \(ax^2+bx+c=0\) (or in \(y\)).
3. Solve the quadratic and back-substitute for the second coordinate.
4. Classify using \(\Delta=b^2-4ac\): \(\Delta>0\) two points; \(\Delta=0\) tangent; \(\Delta<0\) no real intersection.

Method 2 (classification only, no solving)

Distance from the centre \((a,b)\) to the line \(Ax+By+C=0\): \[ d=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}. \] Compare \(d\) with \(r\): if \(d<r\) two intersections; \(d=r\) tangent; \(d>r\) none.

Tangency condition for \(y=mx+c\) to \((x-a)^2+(y-b)^2=r^2\): \[ |am-b+c|=r\sqrt{m^2+1}. \]

Special case

• Vertical line \(x=k\): substitute \(x=k\) into the circle \(\Rightarrow\) quadratic in \(y\).
• Keep answers in exact (surd) form unless asked to approximate.

Quick worked examples

1. Substitution: Intersections of \(x^2+y^2-4x+6y-12=0\) and \(y=x-1\).
   Substitute \(y=x-1\): \[ x^2+(x-1)^2-4x+6(x-1)-12=0 \;\Rightarrow\; 2x^2-17=0 \;\Rightarrow\; x=\pm\frac{\sqrt{34}}{2}. \] Then \(y=x-1\). Points: \(\left(\tfrac{\sqrt{34}}{2},\,\tfrac{\sqrt{34}}{2}-1\right)\), \(\left(-\tfrac{\sqrt{34}}{2},\,-\tfrac{\sqrt{34}}{2}-1\right)\).
2. Discriminant (classification): \(x^2+y^2=10\) with \(y=3x+1\). \[ x^2+(3x+1)^2=10 \Rightarrow 10x^2+6x-9=0,\quad \Delta=6^2-4(10)(-9)=396>0 \] \(\Rightarrow\) two intersections.
3. Vertical line: \((x-3)^2+(y+2)^2=25\) with \(x=7\). \[ (7-3)^2+(y+2)^2=25 \Rightarrow 16+(y+2)^2=25 \Rightarrow y=1 \text{ or } -5. \] Points: \((7,1)\), \((7,-5)\).

Common errors

• Sign mistakes when completing the square (centre signs flip).
• Forgetting the vertical line case \(x=k\).
• Misreading \(\Delta\) (tangent is \(\Delta=0\), not \(\Delta>0\)).
• Premature rounding instead of leaving surds exact.

Practice (exam-style)

1. Find the intersections of \(y=2x-5\) with \(x^2+y^2-6x+8y-11=0\).
2. Decide 0/1/2 intersections for \((x-2)^2+(y+1)^2=13\) and \(3x+4y=20\) (no solving).
3. Find real \(m\) so that \(y=mx-1\) is tangent to \((x-2)^2+(y-1)^2=5\).