Value of Expression for Point on Circle Equidistant from Intersection Points

To find the value of $(6\alpha + 8\beta)^2$, we proceed step-by-step:

Step 1: Identify the properties of the given circle and line

The given circle is: $C : (x - 4)^2 + (y + 3)^2 = 9$

From this equation:

• Center of the circle, $O = (4, -3)$
• Radius of the circle, $R = 3$

The line intersecting the circle at points $Q$ and $R$ is: $x - y = 4$

Step 2: Find the perpendicular bisector of chord $QR$

Since $P(\alpha, \beta)$ is a point on the circle $C$ such that $PQ = PR$, $P$ must lie on the perpendicular bisector of the chord $QR$.

The perpendicular bisector of any chord of a circle always passes through the center of the circle, $O(4, -3)$.

1. The slope of the line $x - y = 4$ is $m_1 = 1$.
2. The slope of the perpendicular bisector is $m_2 = -1$.

Using the point-slope form for the line passing through $O(4, -3)$ with slope $m_2 = -1$: $y - (-3) = -1(x - 4)$ $y + 3 = -x + 4$ $x + y = 1$

Since $P(\alpha, \beta)$ lies on this perpendicular bisector, we have: $\alpha + \beta = 1 \implies \beta = 1 - \alpha$

Step 3: Find the coordinates of $P(\alpha, \beta)$

Since $P(\alpha, \beta)$ also lies on the circle $C$, its coordinates satisfy the equation of the circle: $(\alpha - 4)^2 + (\beta + 3)^2 = 9$

Substitute $\beta + 3 = (1 - \alpha) + 3 = 4 - \alpha$ into the circle's equation: $(\alpha - 4)^2 + (4 - \alpha)^2 = 9$ $2(\alpha - 4)^2 = 9$ $(\alpha - 4)^2 = \frac{9}{2}$ $\alpha - 4 = \pm \frac{3}{\sqrt{2}}$

Step 4: Evaluate the expression $(6\alpha + 8\beta)^2$

Express $6\alpha + 8\beta$ in terms of $(\alpha - 4)$: $6\alpha + 8\beta = 6\alpha + 8(1 - \alpha)$ $= 8 - 2\alpha$ $= 8 - 2(4 + (\alpha - 4))$ $= -2(\alpha - 4)$

Substitute $\alpha - 4 = \pm \frac{3}{\sqrt{2}}$: $6\alpha + 8\beta = -2 \left(\pm \frac{3}{\sqrt{2}}\right) = \mp \frac{6}{\sqrt{2}}$

Now, square the expression: $(6\alpha + 8\beta)^2 = \left(\mp \frac{6}{\sqrt{2}}\right)^2 = \frac{36}{2} = 18$

Thus, the value of $(6\alpha + 8\beta)^2$ is 18.