Find the coordinates of the centre of \(C\) and its radius.

Write down the equation of \(C\).

Find the coordinates of the second point on \(C\) on the chord through \((1,{\text{ }}2)\) parallel to the tangent at \((3,{\text{ }}4)\).

METHOD 1

attempt to exploit the fact that the normal to a tangent passes through the centre \((a,{\text{ }}b)\)     (M1)

EITHER

equation of normal is \(y - 4 =  - \frac{1}{3}(x - 3)\)     (A1)

obtain \(a + 3b = 15\)     A1

attempt to exploit the fact that a circle has a constant radius:     (M1)

obtain \({(1 - a)^2} + {(2 - b)^2} = {(3 - a)^2} + {(4 - b)^2}\)     A1

leading to \(a + b = 5\)     A1

centre is \((0,{\text{ }}5)\)     (M1)A1

radius \( = \sqrt {{1^2} + {3^2}}  = \sqrt {10} \)     A1

OR

gradient of normal \( =  - \frac{1}{3}\)     A1

general point on normal \( = (3 - 3\lambda ,{\text{ }}4 + \lambda )\)     (M1)A1

this point is equidistant from \((1,{\text{ }}2)\) and \((3,{\text{ }}4)\)     M1

if \(10{\lambda ^2} = {(2 - 3\lambda )^2} + {(2 + \lambda )^2}\)

\(10{\lambda ^2} = 4 - 12\lambda  + 9{\lambda ^2} + 4 + 4\lambda  + {\lambda ^2}\)    A1

\(\lambda  = 1\)    A1

centre is \((0,{\text{ }}5)\)     A1

radius \( = \sqrt {10\lambda }  = \sqrt {10} \)     A1

METHOD 2

attempt to substitute two points in the equation of a circle     (M1)

\({(1 - h)^2} + {(2 - k)^2} = {r^2},{\text{ }}{(3 - h)^2} + {(4 - k)^2} = {r^2}\)    A1

 

Note:     The A1 is for the two LHSs, which may be seen equated.

 

equate or subtract the equations

obtain \(h + k = 5\) or equivalent     A1

attempt to differentiate the circle equation implicitly     (M1)

obtain \(2(x - h) + 2(y - k)\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     A1

 

Note:     Similarly, M1A1 if direct differentiation is used.

 

substitute \((3,{\text{ }}4)\) and gradient \( = 3\) to obtain \(h + 3k = 15\)     A1

obtain centre \( = (0,{\text{ }}5)\)     (M1)A1

radius \( = \sqrt {10} \)     A1

[9 marks]

equation of circle is \({x^2} + {(y - 5)^2} = 10\)     A1

[1 mark]

the equation of the chord is \(3x - y = 1\)     A1

attempt to solve the equation for the chord and that for the circle simultaneously     (M1)

for example \({x^2} + {(3x - 1 - 5)^2} = 10\)     A1

coordinates of the second point are \(\left( {\frac{{13}}{5},{\text{ }}\frac{{34}}{5}} \right)\)     (M1)A1

[5 marks]

This question was usually well done, using a variety of valid approaches.

This question was usually well done, using a variety of valid approaches.

This question was usually well done, using a variety of valid approaches.