3.9 - Tangents to a circle

3.9 - Tangents to a circle

Work out the equation of a tangent by using the radius to its point of contact. You will handle circles with any centre, recognise horizontal and vertical exceptions, and combine a tangent equation with intercepts and an exact area.

The radius determines the tangent's gradient

At (3,4)(3,4) on the circle x2+y2=25x^2+y^2=25, the tangent has equation 3x+4y=253x+4y=25. Why do the point's coordinates become the two coefficients? We will first build tangent equations from gradients, then explain that shortcut at the end.

A tangent needs both a direction and a point. Its contact point is given; its direction comes from the radius, which is perpendicular to it. Use the circle's centre as the radius's starting point, rather than automatically using the origin.

You already know

The radius is perpendicular to the tangent at contact; this circle fact is not on the formulae sheet. Non-zero perpendicular gradients are negative reciprocals. The point-gradient equation y−y1=m(x−x1)y-y_1=m(x-x_1) is on the formulae sheet.

Read the centre from (x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2. Check the proposed contact point lies on the circle, calculate the centre-to-point gradient, then use its negative reciprocal through the contact point. A point off the circle is not a contact point at which this method gives a tangent.

[DIAGRAM: asset_name: A shifted radius fixes a tangent direction; asset_slug: l2fm-circle-tangent-gradient; caption: Use coordinate changes from the actual centre, then take the perpendicular direction; description: Equal-scale axes show the circle with centre C at (2, 1), radius 5 and contact P at (6, 4). The sage radius CP has a dashed gradient triangle with horizontal change 4 and vertical change 3. The blue tangent is 4x plus 3y equals 36 and has an accurate right-angle marker with the radius at P. A side calculation states radius gradient 3 quarters and tangent gradient minus 4 thirds.]
Use coordinate changes from the actual centre, then take the perpendicular direction

Worked example 1: a tangent to a shifted circle

Question. Work out the equation of the tangent to (x−2)2+(y−1)2=25(x-2)^2+(y-1)^2=25 at P(6,4)P(6,4).

Plan. Use the centre C(2,1)C(2,1) to obtain the radius gradient, then construct the perpendicular line through PP.

(6−2)2+(4−1)2=16+9=25P lies on the circlemCP=4−16−2=34gradient of the radiusmtangent=−43negative reciprocaly−4=−43(x−6)pass through the contact point4x+3y=36clear the fraction and collect\begin{aligned} (6-2)^2+(4-1)^2&=16+9=25 && P\text{ lies on the circle}\\ m_{CP}&=\frac{4-1}{6-2}=\frac34 && \text{gradient of the radius}\\ m_{\text{tangent}}&=-\frac43 && \text{negative reciprocal}\\ y-4&=-\frac43(x-6) && \text{pass through the contact point}\\ 4x+3y&=36 && \text{clear the fraction and collect} \end{aligned}

Why this step? The radius uses differences from (2,1)(2,1). The ratio 4/64/6 would describe a line from the origin instead.

Answer. 4x+3y=364x+3y=36, or any equivalent equation.

Check. 4(6)+3(4)=364(6)+3(4)=36, so the tangent contains PP. Its gradient −43-\frac43 multiplies the radius gradient 34\frac34 to give −1-1.

Worked example 2 changes only the contact point, reflecting it across the horizontal line through the centre. The vertical radius change reverses sign.

Worked example 2: a positive tangent gradient

Question. Work out the tangent to (x−2)2+(y−1)2=25(x-2)^2+(y-1)^2=25 at Q(6,−2)Q(6,-2).

Plan. Calculate the new radius gradient and use the perpendicular direction through QQ.

(6−2)2+(−2−1)2=16+9=25Q is on the circlemCQ=−2−16−2=−34signed radius gradientmtangent=43negative reciprocaly+2=43(x−6)use the contact point4x−3y=30rearrange exactly\begin{aligned} (6-2)^2+(-2-1)^2&=16+9=25 && Q\text{ is on the circle}\\ m_{CQ}&=\frac{-2-1}{6-2}=-\frac34 && \text{signed radius gradient}\\ m_{\text{tangent}}&=\frac43 && \text{negative reciprocal}\\ y+2&=\frac43(x-6) && \text{use the contact point}\\ 4x-3y&=30 && \text{rearrange exactly} \end{aligned}

Why this step? A falling radius has a rising perpendicular tangent. Changing only one sign without taking the reciprocal would give the wrong direction.

Answer. 4x−3y=304x-3y=30.

Check. At (6,−2)(6,-2), 24+6=3024+6=30, and −34×43=−1-\frac34\times\frac43=-1.

The centre fixes the radius direction but is not a point on the tangent. Substituting the centre into a tangent equation should not make it true for a circle with positive radius.

Watch out

Wrong: use y−1=−43(x−2)y-1=-\frac43(x-2) in Worked example 1. That line goes through the centre. Right: use y−4=−43(x−6)y-4=-\frac43(x-6), through the contact point. The centre supplies the radius; the contact point locates the tangent.

Work out a tangent to a new shifted circle, showing which two points define its radius.

Horizontal and vertical tangents

When the radius is horizontal or vertical, choose the perpendicular tangent directly instead of trying to divide by zero.

A point directly above or below the centre has a vertical radius, so its tangent is horizontal. A point directly left or right of the centre has a horizontal radius, so its tangent is vertical. In either case, the tangent passes through the contact point's unchanged coordinate.

[DIAGRAM: asset_name: Axis-parallel circle tangents; asset_slug: l2fm-axis-circle-tangents; caption: A vertical radius gives a horizontal tangent; a horizontal radius gives a vertical tangent; description: Two equal-scale panels show the circle with centre C at (3, minus 2) and radius 5. Left: a vertical radius reaches P at (3, 3), and the horizontal tangent is y equals 3. Right: a horizontal radius reaches Q at (8, minus 2), and the vertical tangent is x equals 8. Both contact points have right-angle markers between radius and tangent, and the correct tangent equation is labelled.]
A vertical radius gives a horizontal tangent

Axis-parallel tangents

For a circle with centre (a,b)(a,b) and radius r>0r>0,

top and bottom: y=b+r and y=b−rright and left: x=a+r and x=a−r\begin{aligned} \text{top and bottom: }&y=b+r\text{ and }y=b-r\\ \text{right and left: }&x=a+r\text{ and }x=a-r \end{aligned}

These are tangent lines, with contact points on the corresponding vertical or horizontal diameter.
Not on the formulae sheet: learn it.

The point-gradient formula needs a defined gradient. It works for a horizontal tangent with gradient 00, but a vertical tangent needs an equation x=constantx=\text{constant}.

Worked example 3: a horizontal tangent

Question. A circle has equation (x−3)2+(y+2)2=25(x-3)^2+(y+2)^2=25. Work out the tangent at (3,3)(3,3).

Plan. Compare the centre and contact coordinates, then use their perpendicular directions.

C=(3,−2)read the centreP=(3,3)same horizontal coordinate as CCP is verticalthe height changes by 5tangent: y=3horizontal through P\begin{aligned} C&=(3,-2) && \text{read the centre}\\ P&=(3,3) && \text{same horizontal coordinate as }C\\ CP&\text{ is vertical} && \text{the height changes by }5\\ \text{tangent: }y&=3 && \text{horizontal through }P \end{aligned}

Why this step? The radius has no defined gradient, so no reciprocal calculation is available. Its perpendicular direction is still clear.

Answer. y=3y=3.

Check. Substituting y=3y=3 into the circle gives (x−3)2+25=25(x-3)^2+25=25, so x=3x=3 only. There is exactly one contact point.

Worked example 4 changes only the contact position, moving a quarter-turn from the top of the circle to its right-hand point.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account