Mg9, Mg11 - Circle intersections, chords and tangents | OCR A-Level Mathematics B (MEI)

Use the circle properties in Mg11 to connect geometric reasons with coordinate calculations. Then apply Mg9 to find line–circle intersections and recognise tangency through a repeated root, with exact working where a question requires it.

Diameters and right angles

Can a line that meets a circle at one point be identified by algebra alone? For $x^2+y^2=25$ , compare $3x+4y=25$ with $3x+4y=15$ . They are parallel, but one touches and the other cuts the circle. Circle geometry will explain the difference before we count roots.

Here is a striking starting fact: every triangle whose longest side is a diameter and whose third vertex is on that circle is right-angled. The vertex can wander around the circumference; the angle stays fixed.

You already know

The centre of a diameter is its midpoint. Squared distance between points is $(x_2-x_1)^2+(y_2-y_1)^2$ . For finite non-zero gradients, perpendicular lines satisfy $m_1m_2=-1$ .

The condition “on that circle” is essential: points inside it do not generally make a right angle.

Angle in a semicircle

If $AB$ is a diameter and $P$ is a different point on the same circumference, then $\angle APB=90^\circ$ . A chord that misses the centre is not a diameter.

Join the centre $O$ to $P$ . The two pairs of equal radii form isosceles triangles. With their base angles labelled $\alpha$ and $\beta$ , the large triangle's angles are $\alpha$ , $\beta$ , $\alpha+\beta$ . Thus $2(\alpha+\beta)=180^\circ$ . The other semicircle follows by reflection. The theorem is not in the formula sheet: learn it.

Worked example 1: justify the right angle

Question. $A=(-3,1)$ and $B=(5,1)$ are diameter endpoints. Show that $P=(1,5)$ lies on the circle and determine $\angle APB$ .

Plan. Locate the centre, verify the radius-distance and use the diameter theorem.

\begin{aligned} O&=(1,1),\quad r=4 &&\text{midpoint and half-diameter}\\ OP^2&=(1-1)^2+(5-1)^2=16 &&\text{check the circumference} \end{aligned}

Why this step? Membership of the circle is the missing condition that makes the theorem applicable.

Answer. $P$ lies on the circle and $\angle APB=90^\circ$ .

Check. $AP^2=32$ , $BP^2=32$ , $AB^2=64$ , so Pythagoras agrees.

Keep the diameter and change only the position of $P$ , from the top to the bottom of the circle.

Worked example 2: the other semicircle

Question. For the same diameter, show that $Q=(1,-3)$ lies on the circle and determine $\angle AQB$ .

Plan. Repeat the membership check; the theorem works on either side of the diameter.

\begin{aligned} OQ^2&=(1-1)^2+(-3-1)^2=16 &&\text{same radius squared}\\ \angle AQB&=90^\circ &&\text{angle in a semicircle} \end{aligned}

Why this step? The sign of the vertical displacement does not affect its squared length.

Answer. $Q$ is on the circle and the angle is $90^\circ$ .

Check. $m_{AQ}=-1$ and $m_{BQ}=1$ ; their product is $-1$ .

The diameter's endpoints themselves are excluded as third vertices because they would not form a triangle.

Watch out

Wrong: “The line $AB$ is long, so the angle opposite it is right.” Right: identify a diameter and verify that the third point lies on its circle.

Try another circle with a lower-semicircle point.

Chords and perpendicular bisectors

A chord is a straight segment joining two points on a circle. Its midpoint lies on the perpendicular from the centre.

Centre-perpendicular to a chord

The perpendicular from a circle's centre to a chord bisects the chord. Conversely, the perpendicular bisector of a chord passes through the centre.

If $OM$ is perpendicular to $AB$ , right triangles $OMA$ and $OMB$ share $OM$ and have equal hypotenuses, the radii. Their congruence gives $AM=BM$ . This is why one right triangle is enough.

Length of a chord

For radius $r>0$ and perpendicular centre-distance $0\le d<r$ ,

L=2\sqrt{r^2-d^2}.

At $d=r$ the chord collapses to a tangent point; $d>r$ gives no chord. Not in the formula sheet: learn it.

Distance means the shortest, perpendicular distance. A sloping segment to an arbitrary point of the chord will not do.

Worked example 3: from distance to length

Question. A circle has radius $13$ . A chord is $5$ units from its centre. Find its length.

Plan. Find half the chord using Pythagoras, then double.

\begin{aligned} h^2&=13^2-5^2=144 &&\text{radius is the hypotenuse}\\ h&=12 &&\text{positive half-length}\\ L&=24 &&\text{double} \end{aligned}

Why this step? The perpendicular meets the chord at its midpoint, so the right triangle contains only half of it.

Answer. The chord length is $24$ .

Check. $5^2+12^2=169=13^2$ .

Change only the centre-distance, from $5$ to $12$ ; the radius remains $13$ .

Worked example 4: moving towards the circumference

Question. Find the chord length when the centre-distance is $12$ .

Plan. Use the same right-triangle construction.

\begin{aligned} h^2&=13^2-12^2=25 &&\text{Pythagoras}\\ L&=2\sqrt{25}=10 &&\text{restore both halves} \end{aligned}

Why this step? A chord nearer the circumference is shorter, providing a useful sense-check.

Answer. The chord length is $10$ .

Check. $12^2+5^2=13^2$ , and $10<24$ as expected.

A diameter is the limiting central chord: putting $d=0$ gives $L=2r$ .

Watch out

Wrong: a chord with half-length $5$ has length $5$ . Right: state the full length $10$ . The common error is forgetting the second congruent triangle.

Move a chord in a different circle towards its boundary.

A radius determines the tangent

A tangent is perpendicular to the radius at its point of contact. Once the point is known, you can find its equation without calculus.

[DIAGRAM: asset_name: A bisected chord and a perpendicular tangent; caption: A bisected chord and a perpendicular tangent; asset_slug: h640-chord-tangent; description: Equal-scale circles of radius 5 show chord y equals 3 bisected by OM at 0 comma 3, and a tangent 3x plus 4y equals 25 at P 3 comma 4 perpendicular to radius OP]

On the right, every other point of the perpendicular line is farther from $O$ than $P$ is: the radius is one leg of a right triangle. That explains why the line meets the circle only at $P$ .

Radius and tangent gradients

For a finite non-zero radius gradient $m_r$ ,

m_t=-\frac1{m_r}.

A horizontal radius gives a vertical tangent; a vertical radius gives a horizontal tangent. Not in the formula sheet: learn it.

Begin with the useful translated circle from coordinate geometry; its centre is not the origin.

Worked example 5: a sloping tangent

Question. Determine the tangent at $(4,2)$ to $(x-1)^2+(y+2)^2=25$ .

Plan. Check the point, find the radius gradient, then form the perpendicular through it.

\begin{aligned} (4-1)^2+(2+2)^2&=25 &&\text{membership}\\ m_r&=\frac{2+2}{4-1}=\frac43 &&\text{centre }(1,-2)\\ m_t&=-\frac34 &&\text{negative reciprocal}\\ y-2&=-\frac34(x-4) &&\text{through the contact point} \end{aligned}

Why this step? The slope measured from the origin would be the slope of the wrong radius.

Answer. $3x+4y=20$ .

Check. The line contains $(4,2)$ , and $(4/3)(-3/4)=-1$ .

Change only the point of contact to the rightmost point of the same circle. The gradient calculation now needs a geometric boundary case.

Worked example 6: a vertical tangent

Question. Determine the tangent at $(6,-2)$ to the same circle.

Plan. Recognise the horizontal radius and use its perpendicular direction.

\begin{aligned} (6-1)^2+(-2+2)^2&=25 &&\text{point lies on the circle}\\ y_P-y_O&=0 &&\text{radius is horizontal}\\ x&=6 &&\text{vertical line through }P \end{aligned}

Why this step? An undefined tangent gradient is not a failure of the geometry; it tells you to use a vertical line equation.

Answer. $x=6$ .

Check. Substitution into the circle gives $25+(y+2)^2=25$ , so $y=-2$ is its only intersection.

At the top point $(1,3)$ , the same reasoning gives $y=3$ : a vertical radius has a horizontal tangent.

Watch out

Wrong: a vertical tangent must be $y=6$ . Right: a vertical line fixes $x$ , so the tangent at $(6,-2)$ is $x=6$ .

The next point is on the opposite side of the same circle.

Line–circle intersections and tangency

Substitution gives a second, algebraic explanation of tangency. It also finds intersection coordinates when the line cuts across the circle.

The intersection quadratic

After valid line substitution into a non-degenerate circle, the quadratic discriminant $\Delta=b^2-4ac$ decides the number of distinct real intersections:

\Delta>0:\ 2,\qquad \Delta=0:\ 1,\qquad \Delta<0:\ 0.

The one-point case is a tangent. Not in the formula sheet: learn it.

Use exact algebra to establish a result when MEI asks for detailed reasoning. A plot suggests a count; it does not prove a repeated root.

Worked example 7: the cutting line

Question. Find the intersections of $x^2+y^2=25$ with $3x+4y=15$ .

Plan. Express $y$ in terms of $x$ , substitute and solve the quadratic.

\begin{aligned} y&=\frac{15-3x}{4} &&\text{rearrange the line}\\ 16x^2+(15-3x)^2&=400 &&\text{substitute and clear }16\\ 25x^2-90x-175&=0 &&\text{expand}\\ 5(5x+7)(x-5)&=0 &&\text{factorise} \end{aligned}

Why this step? Multiplying by $16$ clears the squared denominator; multiplying by $4$ would not.

Answer. $x=5$ gives $y=0$ , and $x=-7/5$ gives $y=24/5$ . The points are $(5,0)$ , $(-7/5,24/5)$ .

Check. The latter point gives $49/25+576/25=25$ and $-21/5+96/5=15$ .

Now change only the line's constant from $15$ to $25$ . This is the second line in the opening question.

Worked example 8: the tangent line

Question. Show that $3x+4y=25$ is tangent to $x^2+y^2=25$ , and find the contact point.

Plan. Show that substitution produces one repeated root.

\begin{aligned} 16x^2+(25-3x)^2&=400 &&\text{substitute }y=(25-3x)/4\\ 25x^2-150x+225&=0 &&\text{expand and collect}\\ 25(x-3)^2&=0 &&\text{complete the square}\\ x&=3,\quad y=4 &&\text{one coordinate pair} \end{aligned}

Why this step? The squared factor establishes a single distinct real intersection, which is the condition for tangency.

Answer. The line is tangent at $(3,4)$ .

Check. The radius gradient is $4/3$ ; the line gradient is $-3/4$ , so geometry confirms the algebra.

More generally, substituting $3x+4y=k$ gives $25x^2-6kx+k^2-400=0$ . Its discriminant is $64(625-k^2)$ . Hence $|k|<25$ gives two intersections, $|k|=25$ gives tangency and $|k|>25$ gives none. The negative tangent is real too: the circle has two parallel supporting lines, one on each side. A vertical line can instead be substituted directly for $x$ .

Exam tip

For “Show that” and the detailed-reasoning instruction, display the equation that establishes the given result. Writing only “one calculator root” does not explain why the line is tangent.

Keep the same family but move beyond the tangent.

Quick Recap

The right angle in a semicircle requires a diameter and a third point on the circumference.
The centre-perpendicular bisects a chord, so Pythagoras gives half its length.
A tangent is perpendicular to its radius; treat vertical and horizontal cases geometrically.
Line substitution produces a quadratic: two distinct roots, a repeated root or no real roots match the three intersection cases.

The opening lines behave differently because $3x+4y=25$ gives $25(x-3)^2=0$ , touching at $(3,4)$ , whereas $3x+4y=15$ gives two distinct intersection points.

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Mg9, Mg11 - Circle intersections, chords and tangents

Diameters and right angles

Chords and perpendicular bisectors

A radius determines the tangent

Line–circle intersections and tangency

Quick Recap

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