Easy -1.8 This is a straightforward application of the circle equation formula requiring only calculation of radius using distance formula and substitution into (x-a)²+(y-b)²=r². It's a routine C2 exercise with no problem-solving element, significantly easier than average A-level questions.
The equation of the circle is \((x+1)^2 + (y-7)^2 = r^2\)
M1, A1
M1: for this expression on LHS — allow errors in sign of 1 and 7. A1: correct signs (just LHS)
The radius is \(\sqrt{(-1)^2 + 7^2} = \sqrt{50}\) or \(5\sqrt{2}\) or \(r^2 = 50\)
M1
M1: Pythagoras or substitution to give \(r\) or \(r^2\). Giving this value as diameter is M0
So \((x+1)^2 + (y-7)^2 = 50\) or equivalent
A1 (4)
A1 cao for cartesian equation with numerical values; allow \((\sqrt{50})^2\) or \((5\sqrt{2})^2\) or any exact equivalent. A correct answer with no working earns all four marks
# Question 2:
| Working | Marks | Guidance |
|---------|-------|----------|
| The equation of the circle is $(x+1)^2 + (y-7)^2 = r^2$ | M1, A1 | M1: for this expression on LHS — allow errors in sign of 1 and 7. A1: correct signs (just LHS) |
| The radius is $\sqrt{(-1)^2 + 7^2} = \sqrt{50}$ or $5\sqrt{2}$ or $r^2 = 50$ | M1 | M1: Pythagoras or substitution to give $r$ or $r^2$. Giving this value as diameter is **M0** |
| So $(x+1)^2 + (y-7)^2 = 50$ or equivalent | A1 (4) | A1 cao for cartesian equation with numerical values; allow $(\sqrt{50})^2$ or $(5\sqrt{2})^2$ or any exact equivalent. A correct answer with no working earns all four marks |
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2. A circle $C$ has centre $( - 1,7 )$ and passes through the point $( 0,0 )$. Find an equation for $C$.\\
(4)\\
\hfill \mbox{\textit{Edexcel C2 2012 Q2 [4]}}