CAIE A-Level Mathematics 1.3.4 Lines and Circles

Practise solving line–circle intersections, constructing tangents from radius gradients and applying chord, centre and triangle geometry to coordinates, areas and distances.

Exam points

substitute a line into a circle equation and solve the resulting quadratic for intersections
find a tangent gradient as the negative reciprocal of the radius gradient at contact
use centres, chords and radii to calculate triangle areas, tangent intersections or distances

[Maximum number: 6]

Points A(-2,3), B(3,0) and C(6,5) lie on the circumference of a circle with centre D.

(a)

Hence state the coordinates of D.

[ 1 ]

(b)

The point E lies on the circumference of the circle such that B E is a diameter.

Find an equation of the tangent to the circle at E.

[ 5 ]

\left(\frac{x+3}{2}, \frac{y+0}{2}\right)=(2,4) \text { or } \mathbf{B E}=2 \mathbf{B D}=2\binom{-1}{4}
Or Equation of B E is y=-4(x-3) or y-4=-4(x-2) leading to y=-4 x+12 Substitute equation of B E into circle and form a 3-term quadratic.
M1
Use of mid-point formula, vectors, steps on a diagram
May be seen to find x coordinate at E
(x, y)=(1,8) or $\mathbf{O E}=\binom{3}{0}+\binom{-2}{8}=\binom{1}{8}$
A1
E=(1,8)
Accept without working for both marks SC B2
Gradient of B D, m,=-4 or gradient $A C=\frac{1}{4}=$ gradient of tangent
B1
Or gradient of B E=-4
Equation of tangent is y-8=1 / 4(x-1) OE
M1 A1
For M1, equation through their E or ( 1,8 ) (not,
A, B or C ) and with gradient $\frac{-1}{\text { their }-4}$

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1.3.4—Lines and circles question 1

Question (a)

Question (b)