Equation of a circle and tangents – Year 11 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \((x+2)^2+(y-5)^2=9\) (1 for each bracket pair, 1 for 9; award 2 marks for fully correct) Use \((x-a)^2+(y-b)^2=r^2\) with \(a=-2\), \(b=5\) and \(r^2=9\). Subtracting \(-2\) gives \(x+2\).
   2. \((1+2)^2+(5-5)^2=9+0=9\) (1) so it is on the circle (1) Substitute \(x=1\) and \(y=5\): the left-hand side equals \(r^2=9\).

   A common error in (a) is to write \((x-2)^2\) or to leave the radius as 3 instead of \(3^2=9\). In (b), the point \((1,5)\) is 3 units to the right of the centre on a horizontal line, so it is exactly one radius away.

2. Answer to Question 2 (4 marks)

   1. Gradient of \(OA=\dfrac34\) (1); tangent gradient \(-\dfrac43\) (1) The gradient from \(O(0,0)\) to \(A(4,3)\) is \(\tfrac34\). The tangent is perpendicular to the radius, so its gradient is the negative reciprocal.
   2. \(y-3=-\dfrac43(x-4)\) (1), so \(3y-9=-4x+16\) and \(4x+3y=25\) (1) Multiply by 3 to clear the fraction and rearrange. Check: \(4\times4+3\times3=25\).

   In (b), the tangent must pass through \(A(4,3)\); substitution confirms this. A common error is to use the gradient of the radius, \(\tfrac34\), as the tangent gradient.

3. Answer to Question 3 (4 marks)

   1. \(x^2+(x+1)^2=13\) (1), \(2x^2+2x+1=13\), so \(x^2+x-6=0\) (1) Substitute \(y=x+1\) into the circle equation and expand \((x+1)^2=x^2+2x+1\). Subtract 13 and divide by 2.
   2. \((x+3)(x-2)=0\) (1), so \((2,3)\) and \((-3,-2)\) (1) Find \(y\) from \(y=x+1\). Check: \(4+9=13\) and \(9+4=13\).

   In (b), each \(x\)-value must be paired with its own \(y\)-value: \(x=2\) gives \(y=3\) and \(x=-3\) gives \(y=-2\). Since the quadratic has two different roots, the line cuts the circle twice; a repeated root would mean the line is a tangent.

4. Answer to Question 4 (4 marks)

   1. \(OA\perp TA\) (1), \(TA^2=10^2-6^2=64\) (1), \(TA=\) 8 (1) The radius \(OA=6\) is perpendicular to the tangent, so triangle \(OAT\) has a right angle at \(A\) and \(OT=10\) is the hypotenuse.
   2. \(\sin(ATO)=\dfrac6{10}\), so \(36.9^\circ\) (1) \(OA\) is opposite angle \(ATO\) and \(OT\) is the hypotenuse, so we use sine.

   The point \(A\) does not need to be found to answer (a). Check (b) with cosine: \(\cos(ATO)=\tfrac8{10}\) also gives \(36.9^\circ\), and the triangle has sides 6, 8 and 10.

5. Answer to Question 5 (4 marks)

   1. Centre \((3,-1)\): gradient \(\dfrac{5-(-1)}{5-3}=3\) (1) Use the centre \((3,-1)\) and the point \(P(5,5)\): the change in \(y\) is 6 and the change in \(x\) is 2.
   2. Tangent gradient \(-\dfrac13\) (1); \(y-5=-\dfrac13(x-5)\) (1); \(x+3y=20\) (1) The tangent is perpendicular to the radius. Multiply by 3 and rearrange. Check: \(5+15=20\).

   In (a), a common error is to use the origin as the centre; the centre comes from the brackets, \((3,-1)\). Check (b): the point \(P(5,5)\) satisfies \(x+3y=20\).

Total : 20 marks