Standard +0.3 This is a straightforward coordinate geometry question requiring students to find the circle's centre by completing the square, verify the line intersects the circle, and show the triangle is right-angled (likely by showing CA ⊥ CB using perpendicular gradients or Pythagoras). While it involves multiple steps, each is a standard technique with no novel insight required, making it slightly easier than average.
15 The circle \(x ^ { 2 } + y ^ { 2 } + 2 x - 14 y + 25 = 0\) has its centre at the point \(C\). The line \(7 y = x + 25\) intersects the circle at points A and B .
Prove that triangle ABC is a right-angled triangle.
Simplifies to two roots; allow arithmetic errors \([50x^2 + 50x - 600 = 0]\)
\(y = 3, 4\)
A1
Could be solved by calculator \([x = -4, 3]\)
\(A\) and \(B\) are \((-4, 3)\) and \((3, 4)\)
A1
FT their \(y\) values
Gradients of \(AC\) and \(BC\) are \(\frac{4}{3}\) and \(\frac{3}{-4}\)
M1, A1
Attempt gradient of at least one line; both correct gradients (not OA or OB)
Product of gradients is \(-1\) so lines are perpendicular
So triangle is right-angled
A1
Clear argument based on perpendicular lines
Alternative: \(AB^2 = (-4-3)^2 + (3-4)^2 = 50\)
M1
Attempt length of one side; FT coordinates
\(AC^2 = BC^2 = \text{radius}^2 = 25\)
A1
All three correct lengths
\(AC^2 + BC^2 = 50 = AB^2\), so right-angled by Pythagoras
A1
Clear argument based on Pythagoras or cosine rule
## Question 15:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Circle is $(x+1)^2 + (y-7)^2 = 25$ | M1 | Attempts to complete the square for either $x$ or $y$ terms |
| Centre $C$ is $(-1, 7)$ | A1 | Correct coordinates of centre seen or used; condone incorrect/missing radius |
| Intersection: $(7y-25)^2 + y^2 + 14y - 50 - 14y + 25 = 0$ | M1 | Attempt to solve equations simultaneously |
| $50y^2 - 350y + 600 = 0$ | M1 | Simplifies to two roots; allow arithmetic errors $[50x^2 + 50x - 600 = 0]$ |
| $y = 3, 4$ | A1 | Could be solved by calculator $[x = -4, 3]$ |
| $A$ and $B$ are $(-4, 3)$ and $(3, 4)$ | A1 | FT their $y$ values |
| Gradients of $AC$ and $BC$ are $\frac{4}{3}$ and $\frac{3}{-4}$ | M1, A1 | Attempt gradient of at least one line; both correct gradients (not OA or OB) |
| Product of gradients is $-1$ so lines are perpendicular | | |
| So triangle is right-angled | A1 | Clear argument based on perpendicular lines |
| **Alternative:** $AB^2 = (-4-3)^2 + (3-4)^2 = 50$ | M1 | Attempt length of one side; FT coordinates |
| $AC^2 = BC^2 = \text{radius}^2 = 25$ | A1 | All three correct lengths |
| $AC^2 + BC^2 = 50 = AB^2$, so right-angled by Pythagoras | A1 | Clear argument based on Pythagoras or cosine rule |
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15 The circle $x ^ { 2 } + y ^ { 2 } + 2 x - 14 y + 25 = 0$ has its centre at the point $C$. The line $7 y = x + 25$ intersects the circle at points A and B .
Prove that triangle ABC is a right-angled triangle.
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q15 [9]}}