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Step-by-Step Solution
Step 1: Parameterize the Points on the Circle
The equation of the circle is
x^2 + y^2 = 9 .
A convenient way to represent a diameter of this circle is to take one endpoint
P
at
(3\cos \theta,\, 3\sin \theta)
and the other endpoint
Q
at
(-3\cos \theta,\, -3\sin \theta) .
Both points lie on the given circle and indeed form a diameter.
Step 2: Express the Perpendicular Distances
We want to find the perpendicular distances from
P
and
Q
to the straight line
x + y = 2 .
The distance from a point
(x_0,\,y_0)
to the line
Ax + By + C = 0
is given by
\text{Distance}
=
\dfrac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.
The line
x + y = 2
can be rewritten as
x + y - 2 = 0 ,
so here
A = 1 ,
B = 1 ,
and
C = -2 .
Hence, the distance
\alpha
from
P(3\cos \theta,\, 3\sin \theta)
is:
\alpha
=
\dfrac{|3\cos\theta + 3\sin\theta - 2|}{\sqrt{1^2 + 1^2}}
=
\dfrac{|3(\cos\theta + \sin\theta) - 2|}{\sqrt{2}}.
Similarly, the distance
\beta
from
Q(-3\cos \theta,\, -3\sin \theta)
is:
\beta
=
\dfrac{|-3\cos\theta - 3\sin\theta - 2|}{\sqrt{2}}.
Step 3: Formulate the Product \alpha \beta
We need to find
\alpha \beta .
Observe that
\alpha \beta
=
\dfrac{|3(\cos\theta + \sin\theta) - 2|\;|\,-3(\cos\theta + \sin\theta) - 2|}{2}.
We can simplify by considering the expression inside the absolute values.
Let
S = 3(\cos \theta + \sin \theta).
Then
\alpha
=
\dfrac{|S - 2|}{\sqrt{2}},
\quad
\beta
=
\dfrac{|\,(-S) - 2|}{\sqrt{2}}
=
\dfrac{|-S - 2|}{\sqrt{2}}.
Their product is
\alpha \beta
=
\dfrac{|(S - 2)(-S - 2)|}{2}.
Expand
(S - 2)(-S - 2) :
=
-(S^2 - 4S + 4)
=
-S^2 + 4S - 4.
But since there is an absolute value, we consider:
(S - 2)(-S - 2)
=
-\bigl(S^2 - 4S + 4\bigr).
So
\alpha \beta
=
\dfrac{|-S^2 + 4S - 4|}{2}.
Step 4: Express in Terms of \theta and Simplify
Recall
S = 3(\cos\theta + \sin\theta) .
Therefore,
S^2
=
9 (\cos \theta + \sin \theta)^2
=
9(\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta)
=
9(1 + \sin 2\theta).
Hence,
-S^2 + 4S - 4
=
-9(1 + \sin 2\theta)
+ 4 \cdot 3(\cos\theta + \sin\theta)
- 4.
However, a more direct approach (as often done) is to note that
(3\cos \theta + 3\sin \theta)^2
=
9(\cos^2 \theta + 2 \cos \theta \sin \theta + \sin^2 \theta)
=
9(1 + \sin 2\theta).
Substituting into the earlier step, we get
\alpha \beta
=
\left|
\dfrac{(3(\cos\theta + \sin\theta))^2 - 4}{2}
\right|
=
\left|
\dfrac{9 (1 + \sin 2\theta) - 4}{2}
\right|
=
\left|
\dfrac{5 + 9\sin 2\theta}{2}
\right|.
Step 5: Find the Maximum Value
Since
\sin 2\theta
ranges between
-1
and
1 ,
the expression
5 + 9 \sin 2\theta
attains its maximum when
\sin 2\theta = 1 .
In that case,
5 + 9 \sin 2\theta
=
5 + 9
=
14.
Thus,
\alpha \beta_{\max}
=
\dfrac{14}{2}
=
7.
Step 6: Final Answer
Therefore, the maximum value of
\alpha \beta
is
7.
