Moderate -0.3 This is a straightforward application of standard techniques: verify A is on the circle, find the radius, substitute the line equation into the circle equation to get a quadratic, and solve for the second intersection point. While it requires multiple steps (4-5 marks typical), each step uses routine methods with no conceptual difficulty or novel insight required.
5 Points \(A ( 7,12 )\) and \(B\) lie on a circle with centre \(( - 2,5 )\). The line \(A B\) has equation \(y = - 2 x + 26\).
Find the coordinates of \(B\).
OE FT *their* 130, may use distance \(BC\) rather than circle
\((x+2)^2 + (-2x+21)^2 = 130\)
M1
Substitute \(y = -2x+26\) into a circle equation
\(5x^2 - 80x + 315\ [=0]\) leading to \([5](x-9)(x-7)\)
M1
Factorisation OE must be seen
\(x = 9\)
A1
With or without \(x=7\)
\(y = 8\) OR \((9,8)\)
A1
\(y=8\) or \((9,8)\) only. Both A1s dependent on the first M1
6
## Question 5:
| $r^2 = (7+2)^2 + (12-5)^2$ | B1 | Expect 130, may use $AC$ rather than $r$ |
|---|---|---|
| Equation of circle is $(x+2)^2 + (y-5)^2 = 130$ | B1 FT | OE FT *their* 130, may use distance $BC$ rather than circle |
| $(x+2)^2 + (-2x+21)^2 = 130$ | M1 | Substitute $y = -2x+26$ into a circle equation |
| $5x^2 - 80x + 315\ [=0]$ leading to $[5](x-9)(x-7)$ | M1 | Factorisation OE must be seen |
| $x = 9$ | A1 | With or without $x=7$ |
| $y = 8$ OR $(9,8)$ | A1 | $y=8$ or $(9,8)$ only. Both A1s dependent on the first M1 |
| | **6** | |
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5 Points $A ( 7,12 )$ and $B$ lie on a circle with centre $( - 2,5 )$. The line $A B$ has equation $y = - 2 x + 26$.\\
Find the coordinates of $B$.\\
\hfill \mbox{\textit{CAIE P1 2023 Q5 [6]}}