Question

       

     The diagram shows a circle with centre A passing through the point B. A second circle has centre B
     and passes through A. The tangent at B to the first circle intersects the second circle at C and D.
     The coordinates of A are (−1, 4) and the coordinates of B are (3, 2).
     (a) Find the equation of the tangent CBD.                                                                                                             [2]

     (b) Find an equation of the circle with centre B.                                                                                                   [3]

     (c) Find, by calculation, the x-coordinates of C and D.                                                                                        [3]

Answer/Explanation

Ans

9 (a)  \(m_{AB}=\frac{4-2}{-1-3}=-\frac{1}{2}\)

            Equation of tangent is y – 2 = 2 (x – 3)

9 (b) \(AB^{2}=4^{2}+2^{2}=20\ or \ r^{2}=20\ or \ r=\sqrt{20}\ or \ AB=\sqrt{20}\)

           Equation of circle centre B is \((x-3)^{2}+(y-2)^{2}=20\)

9 (c) \((x-3)^{2}+((2x-6)^{2}=their\ 20\)

          \(5x^{2}-30x+25=0 \ or \ 5(x-3)^{2}=20\)

           \(\left [ (5)(x-5)(x-1) \ or x-3=\pm 2 \right ]\ x=5, 1\)

Question

The points A(7, 1), B(7, 9) and C(1, 9) are on the circumference of a circle.
(a) Find an equation of the circle.
(b) Find an equation for the tangent to the circle at B.

Answer/Explanation

Ans:

(a) Centre of circle is (4,5)
\(r^2=(7-4)^2+(1-5)^2\)
r = 5
Equation is \((x-4)^2+(y-5)^2=25\)
(b) Gradient of radius \(=\frac{9-5}{7-4}=\frac{4}{3}\)
Equation of tangent is \(y-9=-\frac{3}{4}(x-7)\)

Question.

  A circle with centre (5, 2) passes through the point (7, 5).

(a) Find an equation of the circle.

The line y = 5x − 10 intersects the circle at A and B.

(b) Find the exact length of the chord AB.

Answer/Explanation

(a)   \(r^{2}\left [ =\left ( 5-2 \right )^{2} + \left ( 7-5 \right )^{2}\right ] = 13\)

Equation of circle is \(\left ( x – 5 \right )^{2} + \left ( y – 2 \right )^{2} = 13\)

(b)

(x -5)2 + (5x – 10 – 2)2

26x2 – 130x + 156 [=0]

[26] (x-2) (x-3) [=0]

(2, 0), (3, 5)

(AB)2 = (3-2)2 + (5 – 0)2

\(AB = \sqrt{26}\)