A9, A16 - Perpendicular lines, circles and tangents

A9, A16 - Perpendicular lines, circles and tangents

Perpendicular gradients connect straight-line algebra (A9) to circle equations and tangents (A16). You will use circles centred at the origin, find a tangent at a given point and recover a contact point from a tangent line.

Perpendicular gradients (Higher)

A circle has centre (0,0)(0,0), radius 55, and a point P=(3,4)P=(3,4) on its edge. Which straight line just touches it at PP? The radius will give us the tangent's direction, but first we need the gradient of a line at right angles to another.

You already know

For a non-vertical line, y=mx+cy=mx+c has gradient mm. Rearrange to this form before reading the gradient, then use a known point to find cc.

Perpendicular lines meet at a right angle. A line of gradient 2/32/3 is perpendicular to a line of gradient −3/2-3/2: reverse the fraction and change its sign.

Perpendicular gradients

m2=−1m1,m1m2=−1m_2=-\frac{1}{m_1},\qquad m_1m_2=-1

These forms apply when both gradients are finite and non-zero. A horizontal line and a vertical line are perpendicular and need separate treatment.

Not on the formula sheet: learn it.

Here is why the relationship works. A movement (a,b)(a,b) along a line has gradient b/ab/a. Turning that direction through a right angle gives movement (−b,a)(-b,a), with gradient −a/b-a/b. Their product is −1-1, provided a,b≠0a,b\ne0. Equal unit scales make the right angle visible; stretching one axis distorts its appearance.

Worked example 1: a negative perpendicular gradient

Question. Find the line through (4,1)(4,1) perpendicular to 3y=6x+93y=6x+9.

Plan. Rearrange the given line, take the negative reciprocal and substitute the point.

y=2x+3divide the whole equation by threem⊥=−12negative reciprocal of two1=−12(4)+csubstitute the pointc=3add twoy=−12x+3write the required line\begin{aligned} y&=2x+3 && \text{divide the whole equation by three} \\ m_{\perp}&=-\frac12 && \text{negative reciprocal of two} \\ 1&=-\frac12(4)+c && \text{substitute the point} \\ c&=3 && \text{add two} \\ y&=-\frac12x+3 && \text{write the required line} \end{aligned}

Why rearrange first? The coefficient 66 in the original equation is not its gradient; the coefficient of yy must be one.

Answer. y=−12x+3y=-\frac12x+3.

Check. At x=4x=4, the output is 11; the gradient product is 2(−1/2)=−12(-1/2)=-1.

Change only the sign of the given line's gradient: 3y=−6x+93y=-6x+9. Keep the required point fixed.

Worked example 2: a positive perpendicular gradient

Question. Find the line through (4,1)(4,1) perpendicular to 3y=−6x+93y=-6x+9.

Plan. Find the new gradient, reverse its fraction and sign, then determine the intercept.

y=−2x+3divide by threem⊥=−1−2=12negative reciprocal1=12(4)+csubstitute the same pointc=−1subtract twoy=12x−1write the equation\begin{aligned} y&=-2x+3 && \text{divide by three} \\ m_{\perp}&=-\frac1{-2}=\frac12 && \text{negative reciprocal} \\ 1&=\frac12(4)+c && \text{substitute the same point} \\ c&=-1 && \text{subtract two} \\ y&=\frac12x-1 && \text{write the equation} \end{aligned}

Why is the new gradient positive? Taking a negative reciprocal changes the sign of a negative gradient.

Answer. y=12x−1y=\frac12x-1.

Check. 12(4)−1=1\frac12(4)-1=1, and (−2)(1/2)=−1(-2)(1/2)=-1.

A gradient of −4-4 pairs with 1/41/4. For y=−3x+1y=-3x+1, the perpendicular gradient is 1/31/3; through (−6,2)(-6,2), the line is y=x/3+4y=x/3+4. A horizontal line has gradient zero, and a vertical line has undefined gradient. The perpendicular to y=7y=7 through (−2,4)(-2,4) is therefore x=−2x=-2; do not try to divide by zero.

Watch out

Wrong: “Gradients 22 and −2-2 are perpendicular.” Changing only the sign gives product −4-4. Right: use the negative reciprocal, so 22 pairs with −1/2-1/2.

Change the given gradient again and use a new point.

The equation of an origin-centred circle (Higher)

Every point on a circle is the same distance from its centre.

For a circle centred at the origin, a point (x,y)(x,y) makes a right triangle with horizontal and vertical distances ∣x∣|x| and ∣y∣|y|, and hypotenuse rr. Pythagoras gives its equation.

Circle centred at the origin

x2+y2=r2x^2+y^2=r^2

The centre is (0,0)(0,0), with radius r>0r>0. The constant is the square of the radius.

Not on the formula sheet: derive it from Pythagoras' theorem.

The circle x2+y2=25x^2+y^2=25 has radius 55. Its axis intercepts are (5,0),(−5,0),(0,5),(0,−5)(5,0),(-5,0),(0,5),(0,-5). The point (3,4)(3,4) lies on it because 9+16=259+16=25. Changing signs or swapping coordinates preserves the squared sum: one 33–44–55 triangle supplies eight different points on this circle.

Worked example 3: finding a coordinate in quadrant I

Question. A point on x2+y2=25x^2+y^2=25 has x=4x=4 and lies in quadrant I. Find yy.

Plan. Substitute the known coordinate, solve for the square and choose the sign from the quadrant.

42+y2=25substitute x=4y2=25−16=9subtract the known squarey=3quadrant I has positive y\begin{aligned} 4^2+y^2&=25 && \text{substitute }x=4 \\ y^2&=25-16=9 && \text{subtract the known square} \\ y&=3 && \text{quadrant I has positive }y \end{aligned}

Why choose only the positive root? The equation alone permits y=3y=3 or −3-3, but the stated quadrant removes the negative value.

Answer. y=3y=3, giving point (4,3)(4,3).

Check. 42+32=254^2+3^2=25, and both coordinates are positive.

Change only the quadrant from I to IV. The squared equation stays unchanged.

Worked example 4: choosing the other sign

Question. A point on x2+y2=25x^2+y^2=25 has x=4x=4 and lies in quadrant IV. Find yy.

Plan. Solve the same squared equation, then use the changed position.

16+y2=25substitute x=4y2=9subtract sixteeny=−3quadrant IV has negative y\begin{aligned} 16+y^2&=25 && \text{substitute }x=4 \\ y^2&=9 && \text{subtract sixteen} \\ y&=-3 && \text{quadrant IV has negative }y \end{aligned}

Why does squaring not select the sign? Both 33 and −3-3 have square 99; the quadrant supplies the missing information.

Answer. y=−3y=-3, giving (4,−3)(4,-3).

Check. 42+(−3)2=254^2+(-3)^2=25; the point is right of the origin and below the horizontal axis.

If no position is specified, retain both signs. For x2+y2=289x^2+y^2=289, radius 289=17\sqrt{289}=17; with x=8x=8, y2=289−64=225y^2=289-64=225, so y=±15y=\pm15. Quadrant I selects 1515.

A point is inside the circle if x2+y2<r2x^2+y^2<r^2, and outside if x2+y2>r2x^2+y^2>r^2, since these compare squared distances with the squared radius. Thus (2,3)(2,3) is inside the radius-five circle: 22+32=13<252^2+3^2=13<25.

For a sketch, use equal unit scales, mark the four axis intercepts and add points such as (3,4)(3,4) and their reflections before drawing the circle. Stretching one axis would make the circle appear elliptical.

Watch out

Wrong: “x2+y2=169x^2+y^2=169 has radius 169169.” Right: r2=169r^2=169, so r=13r=13. A radius is a positive length, not both square roots.

Apply the same coordinate method to a different circle, keeping the quadrant restriction.

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