June 2022 Paper 2 Q13
13 \(A\,(-2, 5)\) and \(B\,(4, 13)\) are points on a circle.
\(AB\) is a diameter.
Work out the equation of the circle.
Give your answer in the form \(\quad (x - a)^2 + (y - b)^2 = c \quad\) where \(a\), \(b\) and \(c\) are integers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x - 1)^2 + (y - 9)^2 = 25\) | B3 | B2 \((x - 1)^2 + (y - 9)^2 = 5^2\) or (1, 9) and radius = 5 or (1, 9) and radius\(^2 = 5^2\) or (1, 9) and radius\(^2\) = 25 B1 \((x - 1)^2 + (y - 9)^2 = k\) or \((x - \ldots)^2 + (y - \ldots)^2 = 25\) or \((x - \ldots)^2 + (y - \ldots)^2 = 5^2\) or (1, 9) or \(\dfrac{-2 + 4}{2}\) oe and \(\dfrac{5 + 13}{2}\) oe or radius = 5 or radius\(^2 = 5^2\) or radius\(^2\) = 25 |
| Alternative method 2 Uses perpendicular lines where \((x, y)\) is a point on the circle | ||
| \(\dfrac{y - 5}{x - -2} \times \dfrac{y - 13}{x - 4} = -1\) | M1 | oe eg \((y - 5)(y - 13) = -1(x + 2)(x - 4)\) |
| \(y^2 - 18y + 65 + x^2 - 2x - 8 = 0\) | M1dep | oe equation of circle with brackets expanded and fractions eliminated eg \(y^2 - 18y + 65 = -x^2 + 2x + 8\) |
| \((x - 1)^2 + (y - 9)^2 = 25\) | A1 | |
Additional guidance
| \(a = 1 \quad b = 9 \quad c = 25\) implies \((x - 1)^2 + (y - 9)^2 = 25\) | |
| Alt 1 (1, 9) may be implied eg \(x = 1 \quad y = 9\) or 1, 9 | B1 |
| Alt 1 \((x + 3)^2 + (y + 4)^2 = 5^2\) | B1 |
| Alt 1 \((x - 1)^2 + (y - 9)^2 = 5\) (with no indication that radius = 5) | B1 |
| Alt 1 \(r = 5\) | B1 |
| Alt 1 diameter = 10 | B0 |
| \((x - 1)^2 + (y - 9)^2 = 25\) in working lines with brackets expanded on answer line | B2 |