1. **Problem statement:** (a)(i) Given the circle equation $x^2 + y^2 = 49$, find the centre and radius. 2. **Formula and rules:** The general form of a circle centered at $(h,k)$ with radius $r$ is: $$ (x - h)^2 + (y - k)^2 = r^2 $$ If the equation is $x^2 + y^2 = r^2$, the centre is at $(0,0)$. 3. **Solution for (a)(i):** Here, $x^2 + y^2 = 49$ means $r^2 = 49$, so $r = \sqrt{49} = 7$. Centre of circle $k$ is $(0,0)$. Radius of circle $k$ is $7$. 4. **(a)(ii) Verify if point $P(4,-6)$ is outside circle $k$:** Calculate distance from centre to $P$: $$ d = \sqrt{(4-0)^2 + (-6-0)^2} = \sqrt{16 + 36} = \sqrt{52} $$ Since $\sqrt{52} \approx 7.21 > 7$, point $P$ lies outside the circle. 5. **(b)(i) Find centre of circle $c$ with diameter endpoints $A(-2,1)$ and $B(6,7)$:** Centre is midpoint: $$ \left( \frac{-2+6}{2}, \frac{1+7}{2} \right) = (2,4) $$ 6. **(b)(ii) Find radius of circle $c$:** Length of diameter $AB$: $$ AB = \sqrt{(6+2)^2 + (7-1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 $$ Radius $r = \frac{AB}{2} = 5$. 7. **(b)(iii) Write equation of circle $c$:** Using centre $(2,4)$ and radius $5$: $$ (x - 2)^2 + (y - 4)^2 = 5^2 $$ $$ (x - 2)^2 + (y - 4)^2 = 25 $$ 8. **(b)(iv) Find values of $a$ for point $(a,4)$ on circle $c$:** Substitute $y=4$ into equation: $$ (a - 2)^2 + (4 - 4)^2 = 25 $$ $$ (a - 2)^2 = 25 $$ $$ a - 2 = \pm 5 $$ So, $$ a = 2 + 5 = 7 $$ $$ a = 2 - 5 = -3 $$ **Final answers:** (a)(i) Centre: $(0,0)$, Radius: $7$ (a)(ii) Point $P(4,-6)$ is outside circle $k$. (b)(i) Centre of $c$: $(2,4)$ (b)(ii) Radius of $c$: $5$ (b)(iii) Equation of $c$: $(x-2)^2 + (y-4)^2 = 25$ (b)(iv) Possible $a$ values: $7$ and $-3$