Circles: equations, centre and radius, tangents

1 Centre and a point on the circle

Centre \((3,-2)\), passing through \((7,1)\).

Askedthe equationKnowthe centreRouter² = (distance from centre to the point)²

1. \(r^2=(7-3)^2+(1+2)^2=16+9=25\)
2. \((x-3)^2+(y+2)^2=25\)

No square root needed: the equation wants \(r^2\).

2 The two ends of a diameter

\((-1,5)\) and \((7,-1)\) are the ends of a diameter.

Askedthe equationKnowtwo ends of a diameterRoutecentre = midpoint (fact 2); r² = (centre to either end)²

1. Centre \(=\left(\tfrac{-1+7}{2},\ \tfrac{5-1}{2}\right)=(3,\,2)\)
2. \(r^2=(7-3)^2+(-1-2)^2=25\)
3. \((x-3)^2+(y-2)^2=25\)

Never use an end of the diameter as the centre, or the full diameter as the radius: the two classic slips.

3 Three points, two of the chords horizontal and vertical

\(A(-2,1)\), \(B(-2,7)\), \(C(6,7)\) lie on a circle.

Askedthe equationKnow\(AB\) is vertical (\(x=-2\) for both), \(BC\) is horizontal (\(y=7\) for both)Routeeither a right angle → diameter, or two easy bisectors

1. Say why it is a right angle: \(AB\) is vertical and \(BC\) is horizontal, so \(AB\perp BC\) and \(\angle ABC=90°\).
2. Name the property: the angle in a semicircle is 90°, so the side opposite the right angle, \(AC\), is a diameter (fact 3).
3. Centre \(=\) midpoint of \(AC=\left(\tfrac{-2+6}{2},\ \tfrac{1+7}{2}\right)=(2,\,4)\); \(r^2=(6-2)^2+(7-4)^2=25\).
4. Check it a second way (fact 1): the bisector of the vertical chord \(AB\) is horizontal through its midpoint, \(y=4\); the bisector of the horizontal chord \(BC\) is vertical through its midpoint, \(x=2\). They meet at \((2,4)\) ✓.
5. \((x-2)^2+(y-4)^2=25\)

If the right angle is not horizontal-and-vertical, prove it with gradients: \(m_{AB}\times m_{BC}=-1\). Then use fact 3 in the same way.

4 Three points, no right angle

\(P(6,6)\), \(Q(-1,7)\), \(R(5,-1)\) lie on a circle (the figure above).

Askedthe equationKnowthree chords, none perpendicularRoutebisectors of two chords; solve them together for the centre

1. Chord \(PQ\): midpoint \(\left(\tfrac52,\tfrac{13}2\right)\), \(m_{PQ}=\tfrac{7-6}{-1-6}=-\tfrac17\); bisector gradient \(7\): \(y=7x-11\)
2. Chord \(QR\): midpoint \((2,3)\), \(m_{QR}=\tfrac{-1-7}{5+1}=-\tfrac43\); bisector gradient \(\tfrac34\): \(y=\tfrac34x+\tfrac32\)
3. Solve: \(7x-11=\tfrac34x+\tfrac32\Rightarrow\tfrac{25}{4}x=\tfrac{25}{2}\Rightarrow x=2,\ y=3\). Centre \((2,3)\)
4. \(r^2=(6-2)^2+(6-3)^2=25\): \((x-2)^2+(y-3)^2=25\). Check \(R\): \(9+16=25\) ✓

You could also substitute all three points into \(x^2+y^2+2gx+2fy+c=0\) and solve three equations, but that is far more algebra and far more room for slips.

5 Two points and a line the centre lies on

A circle passes through \(A(1,5)\) and \(B(8,4)\); its centre lies on \(y=x-3\).

Askedthe equationKnowone chord AB, and a line through the centreRoutebisector of AB meets the given line at the centre

1. Bisector of \(AB\): midpoint \(\left(\tfrac92,\tfrac92\right)\), \(m_{AB}=-\tfrac17\), so gradient \(7\): \(y=7x-27\)
2. Centre: \(7x-27=x-3\Rightarrow x=4,\ y=1\)
3. \(r^2=(1-4)^2+(5-1)^2=25\): \((x-4)^2+(y-1)^2=25\)

Another route: call the centre \((h,\ h-3)\) and set \(CA^2=CB^2\). The \(h^2\) terms cancel and \(h=4\).

6 Two points and the radius: two circles

A circle of radius 5 passes through \(A(1,5)\) and \(B(8,4)\).

Askedboth possible equationsKnowthe centre is on the bisector of AB, and 5 from ARouteput the centre on the bisector, then use CA² = 25

1. Centre on the bisector \(y=7x-27\): call it \((h,\ 7h-27)\)
2. \(CA^2=25\): \((h-1)^2+(7h-32)^2=25\Rightarrow50h^2-450h+1000=0\Rightarrow h^2-9h+20=0\)
3. \(h=4\) or \(h=5\): centres \((4,1)\) and \((5,8)\)
4. \((x-4)^2+(y-1)^2=25\) and \((x-5)^2+(y-8)^2=25\)

Two answers is right: the two centres are reflections of each other in the chord \(AB\), one on each side.

7 A circle that touches an axis

Centre \((-3,4)\); the circle touches the \(x\)-axis.

Askedthe equationKnowtouching the \(x\)-axis means the radius is the height of the centreRouter = |b| (fact 5)

1. The lowest point is on the \(x\)-axis, straight below the centre: \(r=4\)
2. \((x+3)^2+(y-4)^2=16\)

Touching the \(y\)-axis instead would give \(r=|-3|=3\). Touching both axes needs \(|a|=|b|=r\).

8 The other end of a diameter

Centre \((2,3)\); one end of a diameter is \(A(5,7)\). Find the other end \(B\).

AskedBKnowthe centre is the midpoint of ABRoutestep from A to the centre, then the same step again

1. From \(A\) to \(C\): \(x\) goes down 3, \(y\) goes down 4
2. Same step again: \(B=(2-3,\ 3-4)=(-1,-1)\)
3. Quick form: \(B=2C-A=(4-5,\ 6-7)=(-1,-1)\)

1 Tangent at a known point

\((x-3)^2+(y+2)^2=25\). Find the tangent at \(P(7,1)\).

Askedthe tangent at PKnowthe centre (3, −2) and PRouteradius gradient, then the negative reciprocal through P

1. Gradient of the radius \(CP=\tfrac{1-(-2)}{7-3}=\tfrac34\)
2. Tangent gradient \(-\tfrac43\) (tangent ⟂ radius)
3. \(y-1=-\tfrac43(x-7)\Rightarrow4x+3y=31\). Check \(P\): \(28+3=31\) ✓

2 Normal at a known point

Same circle: the normal at \(P(7,1)\).

Askedthe normal at PKnowthe normal is the line CPRouteline through P and the centre

1. Gradient \(\tfrac34\) (the radius itself)
2. \(y-1=\tfrac34(x-7)\Rightarrow3x-4y=17\). The centre fits: \(9+8=17\) ✓

This is how to find an unknown centre when you are told the tangent at a point: the normal at that point passes through the centre (Example 4).

3 Tangents with only the gradient known

\((x-1)^2+(y+2)^2=20\). Find the tangents with gradient 2.

Askedboth tangentsKnowthe gradient, but not the points of contactRoutethe radius to each point of contact is the normal: gradient −½ through C

1. The radius to a point of contact is ⟂ the tangent, so it has gradient \(-\tfrac12\) and goes through \(C(1,-2)\): \(y+2=-\tfrac12(x-1)\)
2. Where this normal meets the circle: \((x-1)^2+\tfrac14(x-1)^2=20\Rightarrow(x-1)^2=16\Rightarrow x=5\) or \(x=-3\)
3. Points of contact \((5,-4)\) and \((-3,0)\): the two ends of a diameter
4. Tangents: \(y+4=2(x-5)\Rightarrow y=2x-14\), and \(y=2(x+3)\Rightarrow y=2x+6\)

Second route (from Quadratics): substitute \(y=2x+c\) into the circle, set the discriminant to 0, and get \(c^2+8c-84=0\), so \(c=6\) or \(-14\). The normal route gives the points of contact as well, which questions often ask for.

4 Tangents from a point outside the circle

See Lines and circles.

Askedthe gradients of both tangentsKnowa point the tangents pass throughRoutey − y₁ = m(x − x₁) into the circle, discriminant = 0

1. Write the line through the outside point with unknown gradient \(m\)
2. Substitute into the circle; one point of contact means \(b^2-4ac=0\)
3. Solve the quadratic in \(m\): two tangents