Hi all
I would appreciate it if someone could help me solve the
following question:
Find the possible equations of the circles that touch both axis
and passes through the point A (-8,1).
Thanks so much!
Dear
knt,
Question
Find
the possible equations of the circles that touch both axis and
passes through the point A(− 8 , 1)
Steps :
(1) Since the equations of the
circles must touch both axis,
this
meansthat the centre of the circle will
be k units vertically
and horizontallyto
both axes and the radius will be k units.
Since the circle must pass through the point A(−
8, 1),
the centre of the
circle must be in the same quadrant too
ie the 2nd quadrant.
Hence, the centre of the circle will be (− k,
k)
(2) Equation
of a circle is (x - a)^2 + (y - b)^2 = r^2
where
(a, b) is the centre of the circle and r is the
radius.
(3) Since the centre is (− k, k) and
radius is k units and a point (x, y)
on the equation of the circle is point A(− 8, 1),
substitute
a = − k, b = k , r =
k, x = − 8, y = 1
(4) (x - a)^2 + (y - b)^2 =
r^2
(-8 -(k))^2 + (1-k)^2
= k^2
(-8+k)^2 + (1-k)^2 = k^2
64 - 16k + k^2 + 1 -
2k + k^2 - k^2 =0
k^2 - 18k + 65
= 0
(k - 5) ( k - 13) =
0
k = 5 or k =
13
(5) (a) When k = 5, the centre is
(− 5, 5) and radius is 5 units
and the equation of the circle
is
(x - (-5))^2 + (y - 5)^2 = 5^2
(x + 5)^2 + (y -5)^2 = 25
(b) When k = 13, the centre is
(− 13, 13) and radius is 13 units
and the equation of the circle
is
(x -(-13))^2 + (y - 13)^2 = 13^2
(x
+ 13)^2 + (y -13)^2 = 169
Thank you for your kind
attention.
Regards,
ahm97sic