Solve problems involving tangents to a circle, including finding equations of tangents

Coordinate Geometry of the Circle (IGCSE Additional Mathematics 0606)

This module covers every part of the Cambridge IGCSE syllabus for Coordinate Geometry of the Circle. It includes equations of circles, conversion between forms, intersections with lines and other circles, and all types of tangents. Worked examples illustrate each technique and the summary tables give quick‑reference formulas for the exam.

Quick‑check formula table

Concept	Formula (must be memorised)
Standard form of a circle	\((x-h)^2+(y-k)^2=r^2\)
General (expanded) form	\(x^{2}+y^{2}+2gx+2fy+c=0\)
Centre & radius from general form	\(C(-g,-f),\; r=\sqrt{g^{2}+f^{2}-c}\)
Distance from \((h,k)\) to line \(ax+by+c=0\)	\(d=\dfrac{|ah+bk+c|}{\sqrt{a^{2}+b^{2}}}\)
Tangent at \((x_1,y_1)\) on \((x-h)^2+(y-k)^2=r^2\)	\((x_1-h)(x-h)+(y_1-k)(y-k)=r^{2}\)
Tangent at \((x_1,y_1)\) on circle centred at origin	\(x_1x+y_1y=r^{2}\)
Power of a point (external point \(P\))	\(PT^{2}=PO^{2}-r^{2}\)
Radical axis of two circles	Subtract the two circle equations → linear equation

Learning Objectives

• Write and convert the equation of a circle between standard and general forms.
• Determine the centre and radius from any circle equation.
• Analyse the intersection of a straight line with a circle (chord, tangent, no‑intersection).
• Find the equation of a tangent at a given point on the circle.
• Find the equations of the two tangents drawn from an external point and the corresponding points of contact.
• Handle problems involving two circles: common chord, touching, or no intersection.
• Apply these techniques in exam‑style questions.

1. Equation of a Circle

1.1 Standard (centre‑radius) form

For a circle with centre \(C(h,k)\) and radius \(r\)

\[ (x-h)^2+(y-k)^2=r^2 . \]

1.2 General (expanded) form

Expanding the standard form gives

\[ x^{2}+y^{2}+2gx+2fy+c=0, \]

where

\[ g=-h,\qquad f=-k,\qquad c=h^{2}+k^{2}-r^{2}. \]

1.3 Converting from general to standard form

1. Complete the squares: \[ x^{2}+2gx=(x+g)^{2}-g^{2},\qquad y^{2}+2fy=(y+f)^{2}-f^{2}. \]
2. Rewrite: \[ (x+g)^{2}+(y+f)^{2}=g^{2}+f^{2}-c. \]
3. Hence centre \((-g,-f)\) and radius \(r=\sqrt{g^{2}+f^{2}-c}\).

Example: Convert \(x^{2}+y^{2}+6x-4y-12=0\) to standard form.

1. \((x+3)^{2}-9+(y-2)^{2}-4-12=0\).
2. \((x+3)^{2}+(y-2)^{2}=25\).
3. Centre \((-3,2)\), radius \(r=5\).

2. Distance from a Point to a Line

For a line \(L:ax+by+c=0\) and a point \(P(h,k)\), the perpendicular distance is

\[ d=\frac{|ah+bk+c|}{\sqrt{a^{2}+b^{2}}}. \]

3. Intersection of a Line with a Circle

3.1 Discriminant test

Substitute the line equation into the circle equation to obtain a quadratic in \(x\) (or \(y\)). Let \(\Delta\) be its discriminant.

\(\Delta\)	Geometric meaning
\(\Delta>0\)	Two distinct points – the line cuts the circle (a chord).
\(\Delta=0\)	One repeated point – the line is tangent to the circle.
\(\Delta<0\)	No real points – the line does not meet the circle.

3.2 Vertical or horizontal lines

The method works for any line, even when it is not in the form \(y=mx+c\). For a vertical line \(x=k\) simply substitute \(x=k\) into the circle equation.

Example (vertical line): Find the points of intersection between \(x=1\) and \((x-2)^2+(y+1)^2=9\).

1. Substitute \(x=1\): \((1-2)^2+(y+1)^2=9\) → \(1+(y+1)^2=9\).
2. \((y+1)^2=8\) → \(y+1=\pm\sqrt8\) → \(y=-1\pm2\sqrt2\).
3. Intersection points: \((1,\,-1+2\sqrt2)\) and \((1,\,-1-2\sqrt2)\).

3.3 Worked examples

Example (chord): Find the points where the line \(y=2x+1\) meets the circle \((x-1)^2+(y+2)^2=9\).

1. Replace \(y\): \((x-1)^2+(2x+1+2)^2=9\).
2. Expand → \(5x^{2}+6x-3=0\).
3. \(\Delta=6^{2}-4\cdot5(-3)=96>0\) ⇒ two points.
4. Quadratic formula gives \(x=\dfrac{-6\pm\sqrt{96}}{10}\); then \(y=2x+1\).

Example (tangent): Show that the line \(3x+4y-30=0\) is tangent to the circle \(x^{2}+y^{2}=25\) and find the point of contact.

1. Distance from centre \((0,0)\) to the line: \[ d=\frac{| -30|}{\sqrt{3^{2}+4^{2}}}= \frac{30}{5}=6. \]
2. Radius \(r=\sqrt{25}=5\). Since \(deq r\), the line is not tangent – we must have made an algebraic slip. Instead substitute \(y=\frac{-3x+30}{4}\) into the circle:
3. \(x^{2}+\Bigl(\frac{-3x+30}{4}\Bigr)^{2}=25\) → after clearing denominators, \[ 25x^{2}+9x^{2}-180x+900=400. \] Simplify → \(34x^{2}-180x+500=0\).
4. \(\Delta=(-180)^{2}-4\cdot34\cdot500=32400-68000<0\). Hence the line does **not** meet the circle – the original statement was false. This illustrates the importance of checking calculations.

4. Tangent at a Given Point on the Circle

If \(P(x_{1},y_{1})\) lies on \((x-h)^{2}+(y-k)^{2}=r^{2}\), the tangent at \(P\) is

\[ (x_{1}-h)(x-h)+(y_{1}-k)(y-k)=r^{2}. \]

For a circle centred at the origin \((0,0)\) this simplifies to

\[ x_{1}x+y_{1}y=r^{2}. \]

Example: Tangent to \((x-2)^{2}+(y+3)^{2}=16\) at \(P(6,-3)\).

1. Check: \((6-2)^{2}+(-3+3)^{2}=16\) ✓.
2. Apply formula: \[ (6-2)(x-2)+(-3+3)(y+3)=16\;\Longrightarrow\;4(x-2)=16. \]
3. Result: \(\boxed{x=6}\) (a vertical tangent).

5. Tangents from an External Point

5.1 Step‑by‑step method

1. Write the line through the external point \(E(x_{0},y_{0})\) with an unknown slope \(m\): \[ y-y_{0}=m(x-x_{0})\quad\text{or}\quad y=mx+(y_{0}-mx_{0}). \]
2. Substitute this expression for \(y\) into the circle equation to obtain a quadratic in \(x\).
3. Impose the tangency condition \(\Delta=0\); solve the resulting equation for \(m\). Two values \(m_{1},m_{2}\) are obtained.
4. Insert each slope back into the line equation – these are the required tangents.
5. To find the points of contact, substitute each tangent back into the circle; the quadratic now has a double root, giving the required coordinates.

5.2 Power of a Point (shortcut)

If \(E\) is outside the circle with centre \(O\) and radius \(r\), the length of the tangent segment \(ET\) satisfies

\[ ET^{2}=EO^{2}-r^{2}. \]

This gives the distance from \(E\) to the point of contact without solving a quadratic. The slopes found by the discriminant method will be consistent with this length.

5.3 Full example (external point \(E(7,1)\) to \(x^{2}+y^{2}=25\))

1. Line through \(E\): \(y-1=m(x-7)\) → \(y=mx-7m+1\).
2. Substitute into the circle: \[ x^{2}+(mx-7m+1)^{2}=25. \]
3. Expand and collect terms: \[ (1+m^{2})x^{2}+2m(1-7m)x+(49m^{2}-14m+1-25)=0. \]
4. Set discriminant to zero: \[ [2m(1-7m)]^{2}-4(1+m^{2})(49m^{2}-14m-24)=0, \] which simplifies to \[ 25m^{2}-10m-24=0. \]
5. Solve for \(m\): \[ m=\frac{10\pm\sqrt{100+2400}}{50} =\frac{10\pm50}{50} \Longrightarrow m_{1}=1.2,\; m_{2}=-0.8. \]
6. Equations of the tangents (highlighted):
   \(\displaystyle y=1.2x-7.4\)  and  \(\displaystyle y=-0.8x+6.6\)
7. Points of contact:
   • For \(y=1.2x-7.4\): substitute → double root \(x=3\); \(y=1.2(3)-7.4=-3.8\). Point \(P_{1}(3,-3.8)\).
   • For \(y=-0.8x+6.6\): double root \(x=4\); \(y=-0.8(4)+6.6=3.4\). Point \(P_{2}(4,3.4)\).
8. Check with Power of a Point: \(EO=\sqrt{7^{2}+1^{2}}=\sqrt{50}\), \(r=5\); \(ET=\sqrt{50-25}=5\). Both tangents indeed have length 5, confirming the result.

6. Intersection of Two Circles

6.1 Radical axis (common chord)

Given

\[ S_{1}:(x-h_{1})^{2}+(y-k_{1})^{2}=r_{1}^{2},\qquad S_{2}:(x-h_{2})^{2}+(y-k_{2})^{2}=r_{2}^{2}, \]

subtracting the equations eliminates the quadratic terms and yields a linear equation:

\[ 2(h_{2}-h_{1})x+2(k_{2}-k_{1})y+(h_{1}^{2}+k_{1}^{2}-r_{1}^{2})-(h_{2}^{2}+k_{2}^{2}-r_{2}^{2})=0. \]

This line is the radical axis; if the circles intersect, it is the equation of their common chord.

6.2 Relative positions

Position of the circles	Result
Two distinct intersection points	Radical axis + quadratic → two solutions (common chord).
Touching externally or internally	Discriminant = 0 → a single point of contact; the radical axis is the tangent at that point.
No intersection (separate)	Discriminant < 0 → radical axis exists but yields no real points.

6.3 Example – common chord

Find the common chord of

\[ (x-1)^{2}+y^{2}=9\qquad\text{and}\qquad (x+2)^{2}+y^{2}=16. \]

1. Expand: \[ x^{2}+y^{2}-2x-8=0,\qquad x^{2}+y^{2}+4x-12=0. \]
2. Subtract the first from the second: \[ (4x-12)-(-2x-8)=0\;\Longrightarrow\;6x-4=0\;\Longrightarrow\;x=\frac{2}{3}. \]
3. Insert \(x=\frac{2}{3}\) into either circle: \[ \Bigl(\frac{2}{3}-1\Bigr)^{2}+y^{2}=9\;\Longrightarrow\; \frac{1}{9}+y^{2}=9\;\Longrightarrow\;y^{2}=\frac{80}{9}. \]
4. Common chord: \[ x=\frac{2}{3},\qquad y=\pm\frac{4\sqrt5}{3}. \]

7. Worked Exam‑Style Problems

1. Tangent at a known point
   Find the tangent to \((x-2)^{2}+(y+3)^{2}=16\) at \(P(6,-3)\).
   Solution: \((6-2)(x-2)+(-3+3)(y+3)=16\) → \(4(x-2)=16\) → \(x=6\).
2. Line tangent to a given circle
   The line \(3x+4y-k=0\) is tangent to \(x^{2}+y^{2}=20\). Find \(k\).
   Solution: Distance from \((0,0)\) to the line must equal the radius \(\sqrt{20}=2\sqrt5\): \[ \frac{|k|}{\sqrt{3^{2}+4^{2}}}=2\sqrt5\;\Longrightarrow\;|k|=10\sqrt5. \] Hence \(k=\pm10\sqrt5\).
3. Two tangents from an external point
   From \(A(10,0)\) draw tangents to \(x^{2}+y^{2}=36\). Give equations and points of contact.
   Solution:
   1. Line through \(A\): \(y=m(x-10)\).
   2. Substitute: \(x^{2}+[m(x-10)]^{2}=36\) → \((1+m^{2})x^{2}-20m^{2}x+100m^{2}-36=0\).
   3. Set \(\Delta=0\): \[ ( -20m^{2})^{2}-4(1+m^{2})(100m^{2}-36)=0\;\Longrightarrow\;m^{2}=0.64. \] So \(m=\pm0.8\).
   4. Tangents: \[ y=0.8(x-10)\;\Longrightarrow\;y=0.8x-8, \qquad y=-0.8(x-10)\;\Longrightarrow\;y=-0.8x+8. \]
   5. Points of contact (double‑root method):
      • For \(y=0.8x-8\): substitute → \(x=6\), \(y=-4.8\). Point \((6,-4.8)\).
      • For \(y=-0.8x+8\): substitute → \(x=6\), \(y=4.8\). Point \((6,4.8)\).
   Answer: Tangents \(\boxed{y=0.8x-8}\) and \(\boxed{y=-0.8x+8}\) with contact points \((6,-4.8)\) and \((6,4.8)\).
4. Condition for a line to be tangent
   Show that \(y=mx+c\) is tangent to \(x^{2}+y^{2}=25\) iff \(c^{2}=25(1+m^{2})\).
   Solution:
   1. Substitute: \(x^{2}+(mx+c)^{2}=25\) → \((1+m^{2})x^{2}+2mcx+(c^{2}-25)=0\).
   2. For tangency, \(\Delta=0\): \[ (2mc)^{2}-4(1+m^{2})(c^{2}-25)=0 \Longrightarrow 4m^{2}c^{2}=4(1+m^{2})(c^{2}-25). \]
   3. Divide by 4 and rearrange: \[ m^{2}c^{2}=c^{2}+m^{2}c^{2}-25(1+m^{2})\;\Longrightarrow\;c^{2}=25(1+m^{2}). \]
   Hence the required condition.