P2 3.1 - Tangents and chords

Tangent

A tangent to a circle is a straight line meeting the circle at exactly one point. The radius to that point is perpendicular to the tangent. For two nonvertical lines with nonzero gradients $m_1,m_2$, perpendicularity means $m_1m_2=-1$; horizontal and vertical lines form the exceptional pair.

Worked example 1: a sloping tangent

Question. Find the tangent at $T(4,6)$ to $(x-1)^2+(y-2)^2=25$.

Plan. Find the radius gradient from $C(1,2)$, take its negative reciprocal, then use the point $T$.

Multiplying out gives $4y-24=-3x+12$.

Answer. $3x+4y=36$.

Check. $T$ satisfies the line, since $3(4)+4(6)=36$. The product of gradients is $(4/3)(-3/4)=-1$, so the line is perpendicular to the radius at a point on the circle.

Worked example 2: a vertical tangent

Question. Find the tangent at $U(6,2)$ to $(x-1)^2+(y-2)^2=25$.

Plan. Recognise the horizontal radius from $(1,2)$ to $(6,2)$, then draw its perpendicular through $U$.

The radius has constant $y=2$. Its perpendicular has constant $x$, and must pass through $U$.

Answer. The tangent is $x=6$.

Check. Substitution into the circle gives $25+(y-2)^2=25$, so $y=2$ is the only intersection. Thus the line touches at exactly $U$.

Watch out

A vertical line has undefined gradient, not gradient zero. Do not try to calculate $-1/0$: use the fact that a horizontal radius has a vertical tangent, and a vertical radius has a horizontal tangent.

Half-chord length

If a chord is a perpendicular distance $d<r$ from the centre of a circle of radius $r$,

The perpendicular meets the chord at its midpoint. A chord through the centre is a diameter and has length $2r$.

Not in the formula booklet: learn it.

Worked example 3: a sloping chord with a known centre

Question. A circle has centre $C(1,2)$ and radius $5$. Its chord endpoints are $A(1,7)$ and $B(29/5,17/5)$. Find the chord length using the perpendicular from the centre.

Plan. Find the chord's midpoint, calculate its distance from the centre, then double the half-chord.

The centre perpendicular meets the chord at $M$, so

Answer. The chord length is $6$.

Check. The direct endpoint distance is $\sqrt{(24/5)^2+(-18/5)^2}=6$, agreeing with the half-chord method.

Worked example 4: locating an unknown centre

Question. $A(1,7)$ and $B(29/5,17/5)$ lie on a circle with centre $C(1,k)$. Find $k$ and an equation for the circle.

Plan. Find the perpendicular bisector of $AB$, put $x=1$ to locate the centre, then calculate the squared radius.

The midpoint remains $M(17/5,26/5)$, and

The perpendicular bisector therefore has equation

The centre lies on this line. At $x=1$, $4-3k=-2$, so $k=2$. Using $A$,

Answer. $k=2$, and the circle is $(x-1)^2+(y-2)^2=25$.

Check. $B$'s squared distance from $(1,2)$ is $(24/5)^2+(7/5)^2=25$, so the second given point lies on this circle too.

Watch out

A chord is not automatically a diameter. Use the centre-to-chord perpendicular to justify the right triangle; the triangle formed by the centre and the two chord endpoints is generally not right-angled.

Angle in a semicircle

If $A,B$ are the endpoints of a diameter and $P$ is another point on the circle, then

The right angle is at the circumference, opposite the diameter.

Not in the formula booklet: learn it.

Worked example 5: checking the right angle

Question. A circle has diameter endpoints $A(-2,1)$ and $B(6,1)$. Verify that $P(2,5)$ is on the circle and that $AP\perp BP$.

Plan. Use the midpoint and radius to check membership, then compare the gradients.

The centre is $(2,1)$ and $r=4$, so the circle is $(x-2)^2+(y-1)^2=16$. At $P$, $0^2+4^2=16$. Also

Answer. $P$ is on the circle and $AP\perp BP$, since the gradients multiply to $-1$.

Check. The squared side lengths are $AP^2=32, BP^2=32, AB^2=64$. Pythagoras independently confirms a right angle at $P$.

Worked example 6: an asymmetric triangle

Question. For the same diameter endpoints, verify that $P(22/5,21/5)$ is on the circle and $AP\perp BP$.

Plan. Check the circle equation and calculate each gradient exactly.

The gradients are

Answer. $P$ is on the circle and the lines are perpendicular because $(1/2)(-2)=-1$.

Check. The squared lengths are $AP^2=256/5$ and $BP^2=64/5$. Their sum is $64=AB^2$, agreeing with the right angle.

Watch out

The centre lies on the diameter, but the theorem's right angle is at $P$, not at the centre. A sketch that happens to look symmetric is not the reason the angle is right.

You already know

For a quadratic $ax^2+bx+c=0$, the discriminant $b^2-4ac$ is positive for two distinct real roots, zero for a repeated real root, and negative for no real roots. Use this after substitution; a sketch alone cannot decide exact tangency.

Worked example 7: two intersections and a chord

Question. Find the intersections of $3x+4y=20$ with $x^2+y^2=25$, and hence find the chord length.

Plan. Substitute $y=(20-3x)/4$, solve the quadratic, then use the distance between the intersection points.

Thus $x=0$ or $24/5$; the corresponding points are $(0,5)$ and $(24/5,7/5)$.

Answer. The chord joins those two points and has length

Check. Both points satisfy the line and circle. Their midpoint is $(12/5,16/5)$, distance $4$ from the origin; the half-chord is therefore $\sqrt{25-16}=3$, confirming total length $6$.

Worked example 8: recognising a tangent algebraically

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

Plan. Substitute the line into the circle and look for a repeated root.

The only solution is $x=3$, giving $y=4$.

Answer. The line is tangent at $(3,4)$, since there is exactly one intersection.

Check. The radius from the origin has gradient $4/3$, while the line has gradient $-3/4$. They are perpendicular, and $(3,4)$ satisfies both equations.

Watch out

A root of the substituted quadratic is an $x$-coordinate, not a complete point. Substitute it into the line to obtain the corresponding $y$, keeping the coordinates paired.