WAEC SSCE - Further Mathematics - 2019…

Find the coordinates of the centre of the circle 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0

(-3. $\frac{9}{2}$)

(-1. $\frac{3}{2}$)

(1, - $\frac{3}{2}$) Azịza ziri ezi

(3. -$\frac{9}{2}$)

Akọwa Nkọwa

To find the center of the circle, we need to rewrite the equation in the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ where (h,k) is the center of the circle and r is the radius. Starting with the given equation: 3x$^2$ + 3y$^2$ - 6x + 9y - 5 = 0 We can group the x and y terms together: (3x$^2$ - 6x) + (3y$^2$ + 9y) = 5 Next, we need to complete the square for both x and y. For the x terms, we can factor out a 3 from the first two terms: 3(x$^2$ - 2x) To complete the square, we need to add and subtract ($\frac{2}{2}$)$^2$ = 1 inside the parenthesis: 3(x$^2$ - 2x + 1 - 1) Then, we can simplify this expression: 3((x - 1)$^2$ - 1) = 3(x - 1)$^2$ - 3 For the y terms, we can follow the same process: 3(y$^2$ + 3y) = 3(y$^2$ + 3y + ($\frac{3}{2}$)$^2$ - ($\frac{3}{2}$)$^2$) = 3((y + $\frac{3}{2}$)$^2$ - $\frac{9}{4}$) = 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ Now we can substitute these expressions back into the original equation and simplify: 3(x - 1)$^2$ - 3 + 3(y + $\frac{3}{2}$)$^2$ - $\frac{27}{4}$ = 5 3(x - 1)$^2$ + 3(y + $\frac{3}{2}$)$^2$ = $\frac{49}{4}$ Dividing both sides by 3, we get: (x - 1)$^2$ + (y + $\frac{3}{2}$)$^2$ = ($\frac{7}{2}$)$^2$ Comparing this equation to the standard form: (x - h)$^2$ + (y - k)$^2$ = r$^2$ We can see that the center of the circle is at (1, -$\frac{3}{2}$) and the radius is ($\frac{7}{2}$). Therefore, the answer is (3). (1, -$\frac{3}{2}$)






Pịa ma Jide iji Dọkpụrụ Ya
Pịa Ebe a ka Imechi

WAEC SSCE - Further Mathematics - 2019 (Objective)