The equation of a circle is (x − a)² + (y − 3)² = 20. The line y = (1/2)x + 6 is a tangent to the circle at the point P.
- (a)Show that one possible value of a is 4 and find the other possible value.[5]
- (b)For a = 4, find the equation of the normal to the circle at P.[4]
- (c)For a = 4, find the equations of the two tangents to the circle which are parallel to the normal found in (b).[4]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [*M1] (x-a)² + (1/2 x+6-3)² = 20 or using x = 2y-12 Obtaining an unsimplified equation in x or y only.
- [A1] 5/4 x² + (3-2a)x + a²-11[=0] OE e.g. 5x² +4(3-2a)x+4a²-44 Rearranging to get a correct 3-term quadratic on one side. Condone terms not grouped together. 5y² - y(54+4a) +133 + a² + 24.
- [DM1] (3-2a)² - 4(5/4)(a²-11)[=0] OE Using b² – 4ac on their 3 term quadratic [= 0].
- [A1] Clearly substituting a=4.
- [B1] a=-16 Condone no method shown for this value.
- [B1] Centre (4, 3) identified or used or the point P is (2, 7)
- [B1] gradient of normal = -2 SOI
- [M1] Forming normal equation using their gradient (not 0.5) and their centre or P Condone use of (±4,±3).
- [A1] (y-3)/(x-4)=-2 or y-7=-2(x-2) OE Condone f(x)=.
- [*M1A1] Coordinates of points at which tangents meet curve are (4+4, 3+2) = (8,5) and (4 – 4, 3 – 2) = (0, 1) Vector approach using their centre and gradient = 0.5. Condone answers only with no working.
- [DM1A1] Equations are y–5=-2(x−8) and y-1=-2x Forming equations of tangents using their (0, 1) and (8,5).
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