PRELIMINARY CONTEST 2024
Contest 34
T.I. AMASS, Kumasi – 71 pts (Winner)
Amenfiman SHS –40 pts
Tuobodom SHTS – 21 pts
Contest 35
Accra Girls’ SHS – 39 pts (Winner)
Notre Dame Girls’ SHS – 34 pts
Aburi Girls’ SHS –27 pts
Contest 36
Anlo SHS – 69 pts (Winner)
Tamale Islamic Science SHS –34 pts
Yendi SHS –28 pts
ROUND 1
PREAMBLE
Find in the form $(x-a)^{2}+(y-b)^{2}=r^{2}$ the equation of the circle touching the y-axis and passing through the point
Given $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$
Find center $(a, b)$
$a=\frac{(x_{1}+x_{2})}{2}$
$b=\frac{(y_{1}+y_{2})}{2}$
Find radius $r^{2}=a^{2}$
1.$A(0, 0) B(6, 0)$
SOLUTION
$x_{1}=0$
$y_{1}=0$
$x_{2}=6$
$y_{2}=0$
$a=\frac{0+6}{2}=3 b=\frac{0+0}{2}=0$
$r^{2}=3^{2}$
$r^{2}=9$
$(x-3)^{2}+(y-0)=9$
$ANSWER= (x-3)^{2}+y^{2}=9$
2.$A(0, 4) B(4, 4)$
SOLUTION
$x_{1}=0$
$y_{1}=4$
$x_{2}=4$
$y_{2}=4$
$a=\frac{0+4}{2}=2 b=\frac{4+4}{2}=4$
$r^{2}=2^{2}$
$r^{2}=4$
$(x-2)^{2}+(y-4)^{2}=4$
$ANSWER: (x-2)^{2}+(y-4)^{2}=4$
3. $A(0, 3) B(-6, 3)$
SOLUTION
$x_{1}=0$
$y_{1}=3$
$x_{2}=-6$
$y_{2}=3$
$a=\frac{0-6}{2}=-3 b=\frac{3+3}{2}=3$
$r^{2}=(-3)^{2}$
$r^{2}=9$
$(x--3)^{2}+(y-3)^{2}=9$
$(x+3)^{2}+(y-3)^{2}=9$
$ANSWER: (x+3)^{2}+(y-3)^{2}=9$
PRACTICE QUESTIONS
Find, in the form $(x-a)^2+(y-b)^2=r^2$, the equation of the circle touching the y-axis and passing through the points $A$ and $B$.
1. $A(0,-1)$, $B(8,-1)$
ANSWER: $(x-4)^2+(y+1)^2=16$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0+8}{2}=4\qquad b=\dfrac{-1-1}{2}=-1$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=4^2=16$
$(x-4)^2+(y+1)^2=16$
2. $A(0,-2)$, $B(-6,-2)$
ANSWER: $(x+3)^2+(y+2)^2=9$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0-6}{2}=-3\qquad b=\dfrac{-2-2}{2}=-2$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=-3^2=9$
$(x+3)^2+(y+2)^2=9$
3. $A(0,4)$, $B(8,4)$
ANSWER: $(x-4)^2+(y-4)^2=16$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0+8}{2}=4\qquad b=\dfrac{4+4}{2}=4$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=4^2=16$
$(x-4)^2+(y-4)^2=16$
4. $A(0,3)$, $B(-8,3)$
ANSWER: $(x+4)^2+(y-3)^2=16$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0-8}{2}=-4\qquad b=\dfrac{3+3}{2}=3$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=-4^2=16$
$(x+4)^2+(y-3)^2=16$
5. $A(0,-6)$, $B(-4,-6)$
ANSWER: $(x+2)^2+(y+6)^2=4$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0-4}{2}=-2\qquad b=\dfrac{-6-6}{2}=-6$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=-2^2=4$
$(x+2)^2+(y+6)^2=4$
6. $A(0,-1)$, $B(6,-1)$
ANSWER: $(x-3)^2+(y+1)^2=9$
SOLUTION
Find the centre $(a,b)$:
$a=\dfrac{0+6}{2}=3\qquad b=\dfrac{-1-1}{2}=-1$
Since the circle touches the y-axis, the radius equals $|a|$:
$r^2=3^2=9$
$(x-3)^2+(y+1)^2=9$