Let circle C be the image of x2+y2−2x+4y−4=0 in the line 2x−3y+5=0 and A be the point on C such that OA is parallel to x -axis and A lies on the right hand side of the centre O of C. If B(α,β), with β<4, lies on C such that the length of the arc AB is (1/6)th of the perimeter of C, then β−3α is equal to
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Step-by-Step Solution
Step 1: Find the center and radius of the original circle
First, we need to find the center and radius of the original circle. We can rewrite the equation of the circle in the standard form
(x−h)2+(y−k)2=r2
by completing the square.
Step 2: Complete the square for the original circle
Completing the square for the
x
terms gives
(x−1)2
and for the
y
terms gives
(y+2)2
. We add and subtract the constants to maintain equality. This shows the original circle has center
P(1,−2)
and radius
r=3
.
Step 3: Find the image of the center of the circle
The image of the circle
C
in the line
2x−3y+5=0
will have the same radius as the original circle. Its center O(h, k) will be the reflection of the original center
P(1,−2)
across the line
2x−3y+5=0
.
Step 4: Calculate the reflected center O
Using the formula for the reflection of a point
(x1,y1)
across the line
ax+by+c=0
, which is
ax−x1=by−y1=−2a2+b2ax1+by1+c
, we find the coordinates of the new center
O
. The center of circle
C
is
O(−3,4)
and its radius is
r=3
.
Step 5: Determine point A
Point
A
is on circle
C
,
OA
is parallel to the
x
-axis, and
A
is to the right of the center
O(−3,4)
. This means
A
has the same
y
-coordinate as
O
and its
x
-coordinate is
xO+r
. So,
A=(−3+3,4)=(0,4)
.
Step 6: Determine point B
The length of arc
AB
is
1/6
th of the perimeter of
C
. This implies that the angle subtended by the arc
AB
at the center
O
is
1/6
th of the total angle in a circle, which is
2π
radians or
360∘
. So,
∠AOB=60∘
.
Step 7: Calculate coordinates of B and the final expression
Point
A
is at
(0,4)
and the center
O
is at
(−3,4)
. The vector
OA
is
(3,0)
, which lies along the positive
x
-axis relative to
O
. Since
∠AOB=60∘
and
β<4
(meaning
B
is below the line
y=4
), point
B
is obtained by rotating
A
by
−60∘
around
O
. The coordinates of
B
are
(xO+rcos(−60∘),yO+rsin(−60∘))
. Substituting the values, we get
α=−3+3(1/2)=−3/2
and
β=4+3(−3/2)=4−33/2
. Finally, we calculate
β−3α=(4−33/2)−3(−3/2)=4−33/2+33/2=4
.