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Coordinate circles and perpendicular lines

Use perpendicular directions, midpoint and area conditions, then recover a circle and verify tangency on the original equation.

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Parallel and perpendicular directions

For a non-vertical line, gradient is change in y divided by change in x. Equal gradients give parallel directions; if the intercepts also match, the lines coincide. Vertical lines have equations x = constant and are parallel to other distinct vertical lines, but have no finite gradient.

When both perpendicular lines have finite gradients, m₁m₂ = -1. Thus a line of gradient 1/2 has perpendicular gradient −2. A horizontal line is perpendicular to a vertical one; do not try to represent that pair by dividing by a zero gradient.

A perpendicular bisector needs both conditions

The midpoint of A(x₁,y₁) and B(x₂,y₂) is ((x₁ + x₂)/2,(y₁ + y₂)/2). A perpendicular bisector passes through that midpoint and has direction perpendicular to AB. A line satisfying only one of those conditions is insufficient.

For A(0,0), B(4,2), the midpoint is (2,1) and AB has gradient 1/2. Hence the perpendicular bisector is y − 1 = −2(x − 2), or 2x + y = 5. For example, (0,5) lies on it and is 5 units from each endpoint: √(0² + 5²) = √(4² + 3²) = 5.

Check direction directly when a coordinate difference is zero:

Direction of ABExample endpointsMidpointPerpendicular bisector
vertical(2,−1), (2,5)(2,2)y = 2, horizontal
horizontal(−1,3), (5,3)(2,3)x = 2, vertical

Check your understanding 1 Separate the original line from its bisector

A(−3,−2) and B(−3,8) have midpoint (−3,3). A proposal says their perpendicular bisector is x = −3 because that line passes through the midpoint. Decide whether it is correct, give the correct equation and check one point’s distances to A and B.

Show answer

The proposal is the original vertical line AB, so it fails the perpendicular condition. The bisector is the horizontal line y = 3. For instance, (0,3) has squared distance 9 + 25 = 34 from both endpoints, confirming equal distances. No product of finite gradients is needed when AB is vertical.

A circle fixes distance from its centre

A circle with centre (a,b) and radius r > 0 contains exactly the points at distance r from its centre. Squaring the coordinate distance gives

(x-a)² + (y-b)² = r².

For x² + y²-6x + 4y-12 = 0, first put the constant on the other side. Complete a square in each coordinate:

(x²-6x + 9) + (y² + 4y + 4) = 12 + 9 + 4,

so (x-3)² + (y + 2)² = 25. The centre is (3,−2) and radius 5, rather than the constant 25. Adding terms inside the squares required adding the same terms to the other side.

In the general form x² + y² + 2gx + 2fy + c = 0, this process gives

(x + g)² + (y + f)² = g² + f²-c.

The centre is (−g,−f). A positive right side gives radius square root of (g² + f²-c). Zero gives a single point, rather than a positive-radius circle; a negative value gives no real points. Reading the centre with the signs of the linear coefficients reversed follows from the completed squares, rather than an extra sign rule.

A centre is equidistant from the endpoints of every chord, so it lies on that chord’s perpendicular bisector. The midpoint of a chord is not necessarily the centre. For x² + y² = 25, the chord from (−3,4) to (3,4) has midpoint (0,4), while the centre is (0,0). The chord’s perpendicular bisector x = 0 contains the centre. Only a stated diameter lets its midpoint identify the centre directly.

Check your understanding 2 Complete both squares

Find the centre and radius of x² + y² + 4x-10y + 13 = 0, showing the completed-square form. Explain why the radius is not the right-hand constant.

Show answer

Adding 4 and 25 while moving the constant gives (x + 2)² + (y-5)² = 16. The centre is (−2,5) and radius 4. The constant is r², so taking its positive square root gives the radius. The same additions must appear on both sides of the equation.

A tangent is perpendicular to its radius

A tangent at a circle point P passes through P and is perpendicular to the radius ending there. Both conditions matter. First check that P lies on the stated circle, then find the radius direction and its perpendicular line through P.

For x² + y² = 25, P(3,4) is on the circle because 3² + 4² = 25. Radius OP has gradient 4/3, so the tangent has gradient −3/4. Through P it is

y-4 = -(3/4)(x-3), or 3x + 4y = 25.

The point satisfies both equations, and the product of the two finite gradients is −1.

The tangent meets the circle perpendicular to its radius

A circle of radius five with equal scales on both coordinate axes. A radius runs from the origin to the marked point (3,4). A descending tangent meets that point without cutting through the circle.

Scroll across the figure to read all labels.

A circle of radius five with equal scales on both coordinate axes. A radius runs from the origin to the marked point (3,4). A descending tangent meets that point without cutting through the circle.A circle of radius five with equal scales on both coordinate axes. A radius runs from the origin to the marked point (3,4). A descending tangent meets that point without cutting through the circle.
The circle is x² + y² = 25, with centre O = (0,0) and radius 5. P = (3,4) lies on the circle. The dashed radius OP has gradient 4/3; the dotted tangent 3x + 4y = 25 has gradient −3/4 and passes through P.
Open full-size graphOpen full-size graph
What to noticeThe equations and coordinate checks establish the radius and tangency exactly; sampled points guide the drawing. The visible tangent is cropped by the plotting window and continues beyond it. A line through a circle point still needs a tangency condition.
View figure data
Figure data: The tangent meets the circle perpendicular to its radius
Seriesx coordinate (coordinate units)y coordinate (coordinate units)
Circle x² + y² = 255.00000.0000
Circle x² + y² = 254.99310.2617
Circle x² + y² = 254.97260.5226
Circle x² + y² = 254.93840.7822
Circle x² + y² = 254.89071.0396
Circle x² + y² = 254.82961.2941
Circle x² + y² = 254.75531.5451
Circle x² + y² = 254.66791.7918
Circle x² + y² = 254.56772.0337
Circle x² + y² = 254.45502.2700
Circle x² + y² = 254.33012.5000
Circle x² + y² = 254.19342.7232
Circle x² + y² = 254.04512.9389
Circle x² + y² = 253.88573.1466
Circle x² + y² = 253.71573.3457
Circle x² + y² = 253.53553.5355
Circle x² + y² = 253.34573.7157
Circle x² + y² = 253.14663.8857
Circle x² + y² = 253.00004.0000
Circle x² + y² = 252.93894.0451
Circle x² + y² = 252.72324.1934
Circle x² + y² = 252.50004.3301
Circle x² + y² = 252.27004.4550
Circle x² + y² = 252.03374.5677
Circle x² + y² = 251.79184.6679
Circle x² + y² = 251.54514.7553
Circle x² + y² = 251.29414.8296
Circle x² + y² = 251.03964.8907
Circle x² + y² = 250.78224.9384
Circle x² + y² = 250.52264.9726
Circle x² + y² = 250.26174.9931
Circle x² + y² = 250.00005.0000
Circle x² + y² = 25-0.26174.9931
Circle x² + y² = 25-0.52264.9726
Circle x² + y² = 25-0.78224.9384
Circle x² + y² = 25-1.03964.8907
Circle x² + y² = 25-1.29414.8296
Circle x² + y² = 25-1.54514.7553
Circle x² + y² = 25-1.79184.6679
Circle x² + y² = 25-2.03374.5677
Circle x² + y² = 25-2.27004.4550
Circle x² + y² = 25-2.50004.3301
Circle x² + y² = 25-2.72324.1934
Circle x² + y² = 25-2.93894.0451
Circle x² + y² = 25-3.14663.8857
Circle x² + y² = 25-3.34573.7157
Circle x² + y² = 25-3.53553.5355
Circle x² + y² = 25-3.71573.3457
Circle x² + y² = 25-3.88573.1466
Circle x² + y² = 25-4.04512.9389
Circle x² + y² = 25-4.19342.7232
Circle x² + y² = 25-4.33012.5000
Circle x² + y² = 25-4.45502.2700
Circle x² + y² = 25-4.56772.0337
Circle x² + y² = 25-4.66791.7918
Circle x² + y² = 25-4.75531.5451
Circle x² + y² = 25-4.82961.2941
Circle x² + y² = 25-4.89071.0396
Circle x² + y² = 25-4.93840.7822
Circle x² + y² = 25-4.97260.5226
Circle x² + y² = 25-4.99310.2617
Circle x² + y² = 25-5.00000.0000
Circle x² + y² = 25-4.9931-0.2617
Circle x² + y² = 25-4.9726-0.5226
Circle x² + y² = 25-4.9384-0.7822
Circle x² + y² = 25-4.8907-1.0396
Circle x² + y² = 25-4.8296-1.2941
Circle x² + y² = 25-4.7553-1.5451
Circle x² + y² = 25-4.6679-1.7918
Circle x² + y² = 25-4.5677-2.0337
Circle x² + y² = 25-4.4550-2.2700
Circle x² + y² = 25-4.3301-2.5000
Circle x² + y² = 25-4.1934-2.7232
Circle x² + y² = 25-4.0451-2.9389
Circle x² + y² = 25-3.8857-3.1466
Circle x² + y² = 25-3.7157-3.3457
Circle x² + y² = 25-3.5355-3.5355
Circle x² + y² = 25-3.3457-3.7157
Circle x² + y² = 25-3.1466-3.8857
Circle x² + y² = 25-2.9389-4.0451
Circle x² + y² = 25-2.7232-4.1934
Circle x² + y² = 25-2.5000-4.3301
Circle x² + y² = 25-2.2700-4.4550
Circle x² + y² = 25-2.0337-4.5677
Circle x² + y² = 25-1.7918-4.6679
Circle x² + y² = 25-1.5451-4.7553
Circle x² + y² = 25-1.2941-4.8296
Circle x² + y² = 25-1.0396-4.8907
Circle x² + y² = 25-0.7822-4.9384
Circle x² + y² = 25-0.5226-4.9726
Circle x² + y² = 25-0.2617-4.9931
Circle x² + y² = 250.0000-5.0000
Circle x² + y² = 250.2617-4.9931
Circle x² + y² = 250.5226-4.9726
Circle x² + y² = 250.7822-4.9384
Circle x² + y² = 251.0396-4.8907
Circle x² + y² = 251.2941-4.8296
Circle x² + y² = 251.5451-4.7553
Circle x² + y² = 251.7918-4.6679
Circle x² + y² = 252.0337-4.5677
Circle x² + y² = 252.2700-4.4550
Circle x² + y² = 252.5000-4.3301
Circle x² + y² = 252.7232-4.1934
Circle x² + y² = 252.9389-4.0451
Circle x² + y² = 253.1466-3.8857
Circle x² + y² = 253.3457-3.7157
Circle x² + y² = 253.5355-3.5355
Circle x² + y² = 253.7157-3.3457
Circle x² + y² = 253.8857-3.1466
Circle x² + y² = 254.0451-2.9389
Circle x² + y² = 254.1934-2.7232
Circle x² + y² = 254.3301-2.5000
Circle x² + y² = 254.4550-2.2700
Circle x² + y² = 254.5677-2.0337
Circle x² + y² = 254.6679-1.7918
Circle x² + y² = 254.7553-1.5451
Circle x² + y² = 254.8296-1.2941
Circle x² + y² = 254.8907-1.0396
Circle x² + y² = 254.9384-0.7822
Circle x² + y² = 254.9726-0.5226
Circle x² + y² = 254.9931-0.2617
Circle x² + y² = 255.00000.0000
Radius OP0.00000.0000
Radius OP3.00004.0000
Tangent 3x + 4y = 250.33336.0000
Tangent 3x + 4y = 256.00001.7500
O and P0.00000.0000
O and P3.00004.0000

At P(5,0) on the same circle, the radius is horizontal and the tangent is vertical, x = 5. At P(0,5), the radius is vertical and the tangent is horizontal, y = 5. These directions give tangents without assigning an undefined gradient or dividing by zero.

Coordinate area needs the actual region

A triangle with vertices (0,0), (6,0) and (2,4) has a horizontal base of 6 and perpendicular height 4, giving area 12 square units. An oblique side is not the perpendicular height.

For a composite polygon, follow its boundary first. The region A(0,0)–B(6,0)–C(6,2)–D(2,5)–E(0,5)–A fits inside a 6 by 5 rectangle. The missing upper-right triangle has horizontal width 6 − 2 = 4 and vertical height 5 − 2 = 3.

The ordered boundary determines which area is wanted

A five-sided region is a rectangle with its upper-right corner removed. The dashed outer corner edges run from D(2,5) through F(6,5) to C(6,2), completing the bounding rectangle. All six points are labelled with coordinates.

Scroll across the figure to read all labels.

A five-sided region is a rectangle with its upper-right corner removed. The dashed outer corner edges run from D(2,5) through F(6,5) to C(6,2), completing the bounding rectangle. All six points are labelled with coordinates.A five-sided region is a rectangle with its upper-right corner removed. The dashed outer corner edges run from D(2,5) through F(6,5) to C(6,2), completing the bounding rectangle. All six points are labelled with coordinates.
The solid boundary is A(0,0)–B(6,0)–C(6,2)–D(2,5)–E(0,5)–A. The outer rectangle is 6 by 5 units. Its missing upper-right triangle CDF has horizontal width 4 and vertical height 3; the diagonal CD is shared with the polygon boundary.
Open full-size graphOpen full-size graph
What to noticeThe polygon and its missing corner are different regions. Choose the requested boundary before applying an area formula. Equal coordinate scales preserve the geometry; the unshaded line diagram represents straight boundaries rather than measured data.
View figure data
Figure data: The ordered boundary determines which area is wanted
Seriesx coordinate (coordinate units)y coordinate (coordinate units)
Polygon boundary A–B–C–D–E–A00
Polygon boundary A–B–C–D–E–A60
Polygon boundary A–B–C–D–E–A62
Polygon boundary A–B–C–D–E–A25
Polygon boundary A–B–C–D–E–A05
Polygon boundary A–B–C–D–E–A00
Outer corner edges D–F–C25
Outer corner edges D–F–C65
Outer corner edges D–F–C62

Subtracting that non-overlapping corner gives area 6 × 5 − (1/2)(4)(3) = 24 square units. The same five coordinates joined in another order could describe a different boundary; the point list alone is insufficient.

Check your understanding 3 Keep the points but change the requested boundary

Using the same labelled points in the figure, find the area of triangle ABD, whose boundary is A–B–D–A. How much smaller is it than the five-sided region? State the base and perpendicular height that belong to the triangle rather than using the polygon’s outline.

Show answer

AB is a horizontal base of length 6 and D has perpendicular height 5 above it. Triangle ABD has area (1/2)(6)(5) = 15 square units, which is 24 − 15 = 9 square units smaller than the polygon. Its boundary omits C and E even though those points remain in the figure. Using the same diagram does not mean the same region was requested.

Try it yourself 1

Find the circle and test the proposed tangent

Minimal support

Problem

A circle has diameter endpoints A(−2,1) and B(4,5). Find its equation in completed-square and expanded forms. Check whether P(3,6) lies on it and, if so, find its tangent at P. A proposal says y = 6 is that tangent because it passes through P. Give a decisive mathematical check of the proposal.

Write your answer, then compare it with the solution

Hints

Hint 1

Use the stated diameter before choosing the tangent direction

Distinguish a diameter’s midpoint from the midpoint of an arbitrary chord. Then check the proposed point in the circle before deciding what direction its tangent must have.

Show full solution

  1. Recover the centre and radius

    Method

    Use the midpoint and half the length of the stated diameter.

    Reason

    A diameter passes through the centre, unlike an arbitrary chord.

    Working

    Answer

    The centre is ((−2 + 4)/2, (1 + 5)/2) = (1,3). The diameter has squared length 6² + 4² = 52, so r² = 52/4 = 13 and r = √13. Hence (x-1)² + (y-3)² = 13, or x² + y²-2x-6y-3 = 0.

  2. Check the point and construct its tangent

    Method

    Test P in the circle and use the radius from (1,3) to P.

    Reason

    A tangent at P needs P on the circle and a direction perpendicular to that radius.

    Working

    Answer

    At P, (3 − 1)² + (6 − 3)² = 4 + 9 = 13, so P is on the circle. The radius gradient is 3/2, making the tangent gradient −2/3. Thus y − 6 = −(2/3)(x − 3), or 2x + 3y = 24. P satisfies this line, and (3/2)(−2/3) = −1.

  3. Test the proposed horizontal line

    Method

    Check whether y = 6 touches only at P.

    Reason

    Passing through a circle point is necessary but is not sufficient for tangency.

    Working

    Answer

    Substituting y = 6 into the circle gives (x − 1)² + 9 = 13, so x = −1 or 3. The line meets the circle at both (−1,6) and P(3,6); it is a secant, rather than the tangent. Its horizontal direction is also not perpendicular to the radius of gradient 3/2.

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