Coordinate geometry of circles flashcards

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Circle
The set of all points at a fixed distance (the radius) from a fixed point (the centre).
Chord
A straight line segment joining two points on a circle. A chord through the centre is a diameter.
Tangent to a circle
A straight line that meets the circle at exactly one point (the point of contact), where it is perpendicular to the radius.
Normal to a circle at a point P
The line through P perpendicular to the tangent at P. For a circle it always passes through the centre.
Perpendicular bisector of a chord
The line through the midpoint of the chord at right angles to it. It always passes through the centre of the circle.
Length of a tangent from P
The distance from an outside point P to the point where a tangent from P touches the circle.
Equation of a circle with centre (a, b) and radius r
(x − a)² + (y − b)² = r² (No — learn it)
Centre and radius of x² + y² + 2gx + 2fy + c = 0
Centre (−g, −f), radius √(g² + f² − c), provided g² + f² − c > 0. (No — derive it)
Equation of a circle with centre the origin
x² + y² = r² (No — learn it)
Condition for a line to be a tangent to a circle (after substituting the line into the circle)
Discriminant b² − 4ac = 0 (one repeated root). (No — learn it)
Length of the tangent from an outside point P to a circle with centre C, radius r
PT² = PC² − r² (Pythagoras, right angle at the point of contact). (No — derive it)
Perpendicular gradients
m₁m₂ = −1, so the perpendicular gradient is −1/m. (No — learn it)
Angle in a semicircle
If AB is a diameter and P is any other point on the circle, angle APB = 90°. (No — learn it)
Perpendicular from the centre to a chord
It bisects the chord; half-chord² + (distance from centre)² = r². (No — learn it)
Two tangents from one outside point
They are equal in length; with the two radii they form a kite symmetric about the line from the point to the centre. (No — learn it)
Line and circle: the three cases
Discriminant > 0: two intersection points (chord); = 0: tangent; < 0: no intersection. (No — learn it)
Circle touching both coordinate axes in the first quadrant
Centre (a, a), radius a: (x − a)² + (y − a)² = a². (No — derive it)
Why does the circle equation have r² on the right?
It is Pythagoras: horizontal distance² + vertical distance² = (distance from the centre)².
How do you tell whether a point is inside, on or outside a circle?
Work out (x − a)² + (y − b)² at the point: less than r² inside, equal on, greater outside.
When is x² + y² + 2gx + 2fy + c = 0 not a circle?
When g² + f² − c ≤ 0: after completing the square the right-hand side is zero or negative.
Why do perpendicular bisectors find the centre of a circle through three points?
The centre is equidistant from all three points, so it lies on the perpendicular bisector of every chord; two bisectors meet at the centre.
How does a right angle help find a circle through three points?
If two sides meet at 90°, the third side is a diameter (converse of the angle in a semicircle); its midpoint is the centre.
Why do parallel tangents touch a circle at the ends of a diameter?
Each is perpendicular to its radius, so both radii lie on one line through the centre.
Centre and radius of x² + y² − 4x + 6y − 12 = 0
(x − 2)² + (y + 3)² = 4 + 9 + 12 = 25: centre (2, −3), radius 5.
Circle with centre (3, 1) passing through (1, 5)
r² = (1 − 3)² + (5 − 1)² = 20: (x − 3)² + (y − 1)² = 20.
Circle with diameter from (−1, 2) to (7, 10)
Centre (3, 6); diameter² = 8² + 8² = 128, so r² = 32: (x − 3)² + (y − 6)² = 32.
Tangent to x² + y² = 25 at (3, 4)
Radius gradient 4/3, tangent gradient −3/4: 3x + 4y = 25.
Tangent length from (7, 13) to (x − 2)² + (y − 1)² = 25
PC² = 5² + 12² = 169; PT² = 169 − 25 = 144, so PT = 12.
Values of k for which y = x + k is a tangent to x² + y² = 8
2x² + 2kx + k² − 8 = 0; discriminant 64 − 4k² = 0, so k = 4 or k = −4.
Signs of the centre
(x + 3)² + (y − 2)² = 16 has centre (−3, 2): the signs in the brackets are opposite to the coordinates.
Coefficients of x² and y² are both 3
Divide the whole equation by 3 before completing the square.
Tangent gradient
Use the negative reciprocal of the radius gradient, not the radius gradient itself.
Expanding (mx + c)² after substituting a line
Keep the cross term: (mx + c)² = m²x² + 2mcx + c².
Discriminant equation with a factor of m
Factorise rather than divide by m: m = 0 (a horizontal tangent) may be a genuine answer.
Exact answers
Leave radii and lengths as simplified surds (e.g. √12.5 = 5√2/2) unless a decimal is asked for.

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