Hi bonkysleuth,
The topic of equation of a circle is a new topic introduced in
the new syllabus 2008 of Additional Mathematics.
The equation of the circle was once set in the late 60s and
early 70s. If you need more practices, you can go to Bras Basah
Complex second hand bookshop to look for the old, old ten years
series.
x^2 + y^2 - 4x + 6y - 12 = 0
x^2 + y^2 + 2x(-2) + 2y(3) - 12 = 0
Since the format of the equation of a circle is
x^2 + y^2 + 2gx + 2fy + c = 0
So, g = - 2 , f = 3, c= - 12
and the centre of a circle is ( -g, -f )
So, the centre = (-(-2), -3) = (2, - 3)
so, the radius = square root (g^2 + f^2 - c) = square root(2^2
+(-3)^2 - (-12)) = 5 units
Since the new circle is a reflection in the y axis, then the
centre of the new circle is (-2, -3) and the radius is also 5
units.
So, the equation of the new circle is
(x - a)^2 + (y- b)^2 = r^2
where centre (a, b) and r is radius
Since the new centre is (-2, -3) and radius is 5 units, so the
equation of the new circle is
(x - a)^2 + (y- b)^2 = r^2
(x -(-2))^2 + (y-(-3))^2 = 5^2
(x+2)^2 +(y+3)^2 = 25
x^2 +4x + 4 + y^2 +6y + 9 -25 = 0
x^2 + y^2 + 4x + 6y -12 = 0
Note : It is to
be noted that reflected in the y axis,
The
old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0
The
new equation of the circle is x^2 + y^2 + 4x + 6y - 12 = 0
It
is to be noted that reflected in the x axis,
The
old equation of the circle is x^2 + y^2 - 4x + 6y - 12 = 0
The
new equation of the circle is x^2 + y^2 - 4x - 6y - 12 = 0
A Practice
Question
(a) Show that the line 4y = x - 3 touches the
circle
x^2 + y^2 - 4x - 8y +
3 = 0
(b) The straight line x = 3 intersects the circle
x^2 + y^2 - 8x - 10y - 9 = 0 at the
points P and Q. Find the
equations of tangents at P and Q.
Regards,
ahm97sic