1. **Problem statement:** (a)(i) Given the circle $k$ with equation $$x^2 + y^2 = 49,$$ find the centre and radius. (a)(ii) Verify if point $P(4, -6)$ lies outside circle $k$. (b)(i) Points $A(-2,1)$ and $B(6,7)$ are endpoints of a diameter of circle $c$. Find the centre of circle $c$. (b)(ii) Find the radius of circle $c$. (b)(iii) Write the equation of circle $c$. (b)(iv) Given point $(a,4)$ lies on circle $c$, find the two possible values of $a$. --- 2. **Formulas and rules:** - The general equation of a circle with centre $(h,k)$ and radius $r$ is $$ (x - h)^2 + (y - k)^2 = r^2.$$ - The centre of a circle in the form $x^2 + y^2 = r^2$ is at the origin $(0,0)$. - Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$ - A point $(x,y)$ lies outside the circle if its distance from the centre is greater than the radius. --- 3. **Solution:** **(a)(i) Centre and radius of circle $k$:** - Equation: $$x^2 + y^2 = 49$$ - This matches the form $(x - 0)^2 + (y - 0)^2 = 7^2$. - So, centre is $(0,0)$ and radius is $7$. **(a)(ii) Verify if $P(4,-6)$ is outside circle $k$:** - Calculate distance from centre $(0,0)$ to $P(4,-6)$: $$d = \sqrt{(4 - 0)^2 + (-6 - 0)^2} = \sqrt{16 + 36} = \sqrt{52}.$$ - Since $\sqrt{52} \approx 7.21 > 7$, point $P$ lies outside the circle. **(b)(i) Centre of circle $c$ with diameter endpoints $A(-2,1)$ and $B(6,7)$:** - Centre is midpoint of $A$ and $B$: $$\left( \frac{-2 + 6}{2}, \frac{1 + 7}{2} \right) = (2, 4).$$ **(b)(ii) Radius of circle $c$:** - Diameter length: $$d = \sqrt{(6 - (-2))^2 + (7 - 1)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10.$$ - Radius is half diameter: $$r = \frac{10}{2} = 5.$$ - Show cancellation: $$r = \frac{\cancel{10}}{\cancel{2}} = 5.$$ **(b)(iii) Equation of circle $c$:** - Centre $(2,4)$ and radius $5$: $$ (x - 2)^2 + (y - 4)^2 = 5^2 = 25.$$ **(b)(iv) Find $a$ such that point $(a,4)$ lies on circle $c$:** - Substitute $y=4$ into equation: $$ (a - 2)^2 + (4 - 4)^2 = 25$$ $$ (a - 2)^2 + 0 = 25$$ $$ (a - 2)^2 = 25$$ - Taking square root: $$ a - 2 = \pm 5$$ - So, $$ a = 2 + 5 = 7 \quad \text{or} \quad a = 2 - 5 = -3.$$ --- **Final answers:** - Centre of $k$: $(0,0)$ - Radius of $k$: $7$ - Point $P(4,-6)$ is outside circle $k$. - Centre of $c$: $(2,4)$ - Radius of $c$: $5$ - Equation of $c$: $$(x - 2)^2 + (y - 4)^2 = 25$$ - Possible values of $a$: $7$ and $-3$