If P$P$ is a point on the circle x^2 + y^2 = 4$x^{2} + y^{2} = 4$, Q$Q$ is a point on the straight line 5x + y + 2 = 0$5x + y + 2 = 0$ and x - y + 1 = 0$x - y + 1 = 0$ is the perpendicular bisector of PQ$PQ$, then 13 times the sum of abscissa of all such point P$P$ is ____.

### Related Formula textPerpendicular bisector properties: m_1m_2 = -1 text and mid-point lies on the line.$$\text{Perpendicular bisector properties: } m_1m_2 = -1 \text{ and mid-point lies on the line.}$$ textParametric point on circle x^2+y^2=r^2 text is (rcostheta, rsintheta)$$\text{Parametric point on circle } x^2+y^2=r^2 \text{ is } (r\cos\theta, r\sin\theta)$$ ### Core Logic

Let P = (2costheta, 2sintheta)$P = (2\cos\theta, 2\sin\theta)$. Let Q$Q$ on the line 5x + y + 2 = 0$5x + y + 2 = 0$ be Q(alpha, -5alpha-2)$Q(\alpha, -5\alpha-2)$. The line x - y + 1 = 0$x - y + 1 = 0$ is the perpendicular bisector of PQ$PQ$. This gives two conditions: slope of PQ$PQ$ is -1$-1$, and mid-point of PQ$PQ$ satisfies the bisector equation. ### Step 1: Apply Slope Condition Slope of bisector is 1$1$, so slope of PQ$PQ$ must be -1$-1$. frac2sintheta - (-5alpha - 2)2costheta - alpha = -1$$\frac{2\sin\theta - (-5\alpha - 2)}{2\cos\theta - \alpha} = -1$$ 2sintheta + 5alpha + 2 = -2costheta + alpha$$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$$ sintheta + costheta + 2alpha + 1 = 0 quad dots (1)$$\sin\theta + \cos\theta + 2\alpha + 1 = 0 \quad \dots (1)$$ ### Step 2: Apply Midpoint Condition Midpoint of PQ$PQ$ is left( frac2costheta + alpha2, frac2sintheta - 5alpha - 22 right)$\left( \frac{2\cos\theta + \alpha}{2}, \frac{2\sin\theta - 5\alpha - 2}{2} \right)$. Substitute into x - y + 1 = 0$x - y + 1 = 0$: frac2costheta + alpha2 - frac2sintheta - 5alpha - 22 + 1 = 0$$\frac{2\cos\theta + \alpha}{2} - \frac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$$ 2costheta + alpha - 2sintheta + 5alpha + 2 + 2 = 0$$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$$ costheta - sintheta + 3alpha + 2 = 0 quad dots (2)$$\cos\theta - \sin\theta + 3\alpha + 2 = 0 \quad \dots (2)$$ ### Step 3: Eliminate alpha and Solve From (1), 2alpha = -sintheta - costheta - 1 implies alpha = frac-sintheta - costheta - 12$2\alpha = -\sin\theta - \cos\theta - 1 \implies \alpha = \frac{-\sin\theta - \cos\theta - 1}{2}$. Substitute alpha$\alpha$ into (2): costheta - sintheta + 3left( frac-sintheta - costheta - 12 right) + 2 = 0$$\cos\theta - \sin\theta + 3\left( \frac{-\sin\theta - \cos\theta - 1}{2} \right) + 2 = 0$$ 2costheta - 2sintheta - 3sintheta - 3costheta - 3 + 4 = 0$$2\cos\theta - 2\sin\theta - 3\sin\theta - 3\cos\theta - 3 + 4 = 0$$ -costheta - 5sintheta + 1 = 0 implies costheta + 5sintheta = 1$$-\cos\theta - 5\sin\theta + 1 = 0 \implies \cos\theta + 5\sin\theta = 1$$ Let's express in half angles: 1 - 2sin^2fractheta2 + 10sinfractheta2cosfractheta2 = 1$$1 - 2\sin^2\frac{\theta}{2} + 10\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 1$$ 2sinfractheta2 left( 5cosfractheta2 - sinfractheta2 right) = 0$$2\sin\frac{\theta}{2} \left( 5\cos\frac{\theta}{2} - \sin\frac{\theta}{2} \right) = 0$$ So, sinfractheta2 = 0 implies costheta = 1$\sin\frac{\theta}{2} = 0 \implies \cos\theta = 1$ or tanfractheta2 = 5 implies costheta = frac1 - tan^2(theta/2)1 + tan^2(theta/2) = frac1 - 251 + 25 = -frac2426 = -frac1213$\tan\frac{\theta}{2} = 5 \implies \cos\theta = \frac{1 - \tan^2(\theta/2)}{1 + \tan^2(\theta/2)} = \frac{1 - 25}{1 + 25} = -\frac{24}{26} = -\frac{12}{13}$. ### Step 4: Final Calculation The abscissa of P$P$ is 2costheta$2\cos\theta$. Values of abscissa are 2(1) = 2$2(1) = 2$ and 2left(-frac1213right) = -frac2413$2\left(-\frac{12}{13}\right) = -\frac{24}{13}$. Sum of abscissa values = 2 - frac2413 = frac26 - 2413 = frac213$= 2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$. We need 13 times (textSum) = 13 times frac213 = 2$13 \times (\text{Sum}) = 13 \times \frac{2}{13} = 2$. ### Pattern Recognition Instead of finding the image of a generic circle point in a line, construct the reflection point Q$Q$ parameter, use slope logic (m_1m_2=-1$m_1m_2=-1$) and midpoint logic simultaneously to create a trigonometric linear equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Circles Class 11 Maths: Straight Lines