<section id="p1-line-circle-intersections"> <h2>Lines &amp; Circles — Intersections (9709 P1)</h2> <p><strong>Learning objective:</strong> 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.</p> <h3>Key facts</h3> <ul> <li>Circle (standard form): \((x-a)^2+(y-b)^2=r^2\). Centre \((a,b)\), radius \(r\).</li> <li>Circle (general form): \(x^2+y^2+gx+fy+c=0\). Centre \(\left(-\tfrac{g}{2},-\tfrac{f}{2}\right)\), radius \(r=\sqrt{\left(\tfrac{g}{2}\right)^2+\left(\tfrac{f}{2}\right)^2-c}\).</li> <li>Line forms: \(y=mx+c\), \(Ax+By+C=0\), and vertical \(x=k\).</li> <li>Graph–equation link: intersection points of the graphs \(\Leftrightarrow\) simultaneous solutions of the equations.</li> </ul> <h3>Method 1 (coordinates via substitution)</h3> <ol> <li>Write the line as \(y=mx+c\) (or handle \(x=k\) separately).</li> <li>Substitute into the circle to get a quadratic \(ax^2+bx+c=0\) (or in \(y\)).</li> <li>Solve the quadratic and back-substitute for the second coordinate.</li> <li>Classify using \(\Delta=b^2-4ac\): \(\Delta&gt;0\) two points; \(\Delta=0\) tangent; \(\Delta&lt;0\) no real intersection.</li> </ol> <h3>Method 2 (classification only, no solving)</h3> <p>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 \(d&lt;r\) two intersections; \(d=r\) tangent; \(d&gt;r\) none. </p> <p><em>Tangency condition</em> for \(y=mx+c\) to \((x-a)^2+(y-b)^2=r^2\): \[ |am-b+c|=r\sqrt{m^2+1}. \]</p> <h3>Special case</h3> <ul> <li><strong>Vertical line</strong> \(x=k\): substitute \(x=k\) into the circle \(\Rightarrow\) quadratic in \(y\).</li> <li>Keep answers in exact (surd) form unless asked to approximate.</li> </ul> <h3>Quick worked examples</h3> <ol> <li><strong>Substitution</strong>: Intersections of \(x^2+y^2-4x+6y-12=0\) and \(y=x-1\). <br>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: \(\left(\tfrac{\sqrt{34}}{2},\,\tfrac{\sqrt{34}}{2}-1\right)\), \(\left(-\tfrac{\sqrt{34}}{2},\,-\tfrac{\sqrt{34}}{2}-1\right)\).</li> <li><strong>Discriminant (classification)</strong>: \(x^2+y^2=10\) with \(y=3x+1\). \[ x^2+(3x+1)^2=10 \Rightarrow 10x^2+6x-9=0,\quad \Delta=6^2-4(10)(-9)=396&gt;0 \] \(\Rightarrow\) two intersections.</li> <li><strong>Vertical line</strong>: \((x-3)^2+(y+2)^2=25\) with \(x=7\). \[ (7-3)^2+(y+2)^2=25 \Rightarrow 16+(y+2)^2=25 \Rightarrow y=1 \text{ or } -5. \] Points: \((7,1)\), \((7,-5)\).</li> </ol> <h3>Common errors</h3> <ul> <li>Sign mistakes when completing the square (centre signs flip).</li> <li>Forgetting the vertical line case \(x=k\).</li> <li>Misreading \(\Delta\) (tangent is \(\Delta=0\), not \(\Delta&gt;0\)).</li> <li>Premature rounding instead of leaving surds exact.</li> </ul> <h3>Practice (exam-style)</h3> <ol> <li>Find the intersections of \(y=2x-5\) with \(x^2+y^2-6x+8y-11=0\).</li> <li>Decide 0/1/2 intersections for \((x-2)^2+(y+1)^2=13\) and \(3x+4y=20\) (no solving).</li> <li>Find real \(m\) so that \(y=mx-1\) is tangent to \((x-2)^2+(y-1)^2=5\).</li> </ol> <details> <summary>Show answers</summary> <p>1) \(\left(1\pm \tfrac{\sqrt{155}}{5},\; -3\pm \tfrac{2\sqrt{155}}{5}\right)\). <br>2) Centre \((2,-1)\), \(r=\sqrt{13}\); distance \(=\dfrac{|6-4-20|}{5}=3.6&lt;\sqrt{13}\Rightarrow\) two intersections. <br>3) \(\dfrac{|2m-1-1|}{\sqrt{m^2+1}}=\sqrt{5}\Rightarrow m^2+8m+1=0 \Rightarrow m=-4\pm\sqrt{15}\).</p> </details> </section>