Straight lines and circles – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Find the gradient, then use \(y-y_1=m(x-x_1)\) with \(A(1,3)\). Gradient \(=\dfrac{11-3}{5-1}=2\) (1). \(y-3=2(x-1)\) so \(y=2x+1\) (1) Check with \(B\): \(2\times5+1=11\) ✓.
   2. The perpendicular bisector passes through the midpoint of \(AB\) and has gradient \(-1\div2=-\tfrac12\). Midpoint \((3,7)\) (1). Gradient \(-\tfrac12\): \(y-7=-\tfrac12(x-3)\), so \(y=-\tfrac12x+\tfrac{17}2\) (1) Check: the midpoint \((3,7)\) satisfies \(-\tfrac32+\tfrac{17}2=7\) ✓.

   The perpendicular bisector can also be written \(x+2y=17\), which is often the neater form to quote.

2. Answer to Question 2 (4 marks)

   1. Complete the square in \(x\) and in \(y\), then read off the centre and the radius. \((x-5)^2+(y+2)^2=25+4-13=16\) (1). Centre \((5,-2)\) (1), radius 4 (1) So the circle has centre \((5,-2)\) and radius 4.
   2. Compare the squared distance from the centre with \(r^2=16\). \((1-5)^2+(-1+2)^2=17\gt 16\) so it lies outside (1) The point is just outside the circle.

   A sign slip giving the centre \((-5,2)\) is the most common error.

3. Answer to Question 3 (4 marks)

   1. Substitute \(P\) into the left-hand side. \((7-3)^2+(1+2)^2=16+9=25\) (1) So \(P\) lies on the circle.
   2. The tangent is perpendicular to the radius from the centre \((3,-2)\) to \(P\). Radius gradient \(\tfrac{3}{4}\) (1). Tangent gradient \(-\tfrac43\) (1). \(y-1=-\tfrac43(x-7)\), i.e. \(4x+3y=31\) (1) Check: \(4\times7+3\times1=31\) ✓.

   Radius gradient: \(\dfrac{1-(-2)}{7-3}=\dfrac34\). The product of the gradients is \(\tfrac34\times(-\tfrac43)=-1\) ✓, so the lines are perpendicular. Using the radius gradient itself gives the normal, not the tangent.

4. Answer to Question 4 (3 marks)

   Substitute the line into the circle. A tangent meets the circle at exactly one point, so the resulting quadratic has equal roots and its discriminant is zero.

   \(x^2+(x+k)^2=8\Rightarrow2x^2+2kx+k^2-8=0\) (1). Discriminant \(4k^2-8(k^2-8)=64-4k^2=0\) (1), so \(k=\pm4\) (1).

   So the lines \(y=x+4\) and \(y=x-4\) are the two tangents; they touch the circle at \((-2,2)\) and \((2,-2)\).

   Check: the distance from the origin to \(x-y+4=0\) is \(\dfrac4{\sqrt2}=2\sqrt2=\sqrt8\), the radius ✓.

5. Answer to Question 5 (5 marks)

   1. The centre is the midpoint of the diameter and the radius is half its length. Centre \((6,6)\) (1). \(AB=\sqrt{64+36}=10\) so radius 5 (1). \((x-6)^2+(y-6)^2=25\) (1)
   2. Substitute \(C\) into the equation of the circle. \((9-6)^2+(10-6)^2=9+16=25\) (1) So the kiosk lies on the boundary.
   3. Perpendicular lines have gradients that multiply to \(-1\). Gradient \(AC=\dfrac{7}{7}=1\) and gradient \(BC=\dfrac{1}{-1}=-1\); product \(-1\) (1) So angle \(ACB=90^\circ\), as expected for the angle in a semicircle.

Total : 20 marks