If P is a point on the circle x² + y² = 4, Q is a point on the straight line 5x + y + 2 = 0 and x -

Q: Why did we choose parametric coordinates for $P$ instead of just $(x, y)$?

Great question! Using $(x, y)$ would force you to carry the constraint $x^2 + y^2 = 4$ throughout your algebra, which is messy. Parametric coordinates $(2\cos\theta, 2\sin\theta)$ automatically satisfy the circle equation, effectively reducing your degrees of freedom and making the trigonometric manipulation much cleaner.

Q: Why do we need two separate equations to solve for $\alpha$ and $\theta$?

Think of it as a system of constraints. We have two unknowns, $\alpha$ (the position of $Q$) and $\theta$ (the position of $P$). To lock down these points, we need two independent pieces of information: the midpoint condition (which fixes the 'center' of the segment) and the slope condition (which fixes the 'orientation' of the segment). Without both, the system is underdetermined.

Q: Is there a faster way to solve $\cos\theta + 5\sin\theta = 1$ without half-angle substitution?

You could use the harmonic form $R\cos(\theta - \phi) = 1$, but that would involve inverse trigonometric functions like $\tan^{-1}(5)$, which makes finding the exact values of $\cos\theta$ much harder. Half-angle substitution is the 'gold standard' for JEE because it converts the trigonometric equation into a simple quadratic in $t = \tan(\theta/2)$, leading directly to rational values.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a circle defined by the equation $x^2 + y^2 = 4$ and a line defined by $5x + y + 2 = 0$. We seek a point $P$ on the circle and a point $Q$ on the line such that the line $x - y + 1 = 0$ is the perpendicular bisector of the segment $PQ$.

Since $P$ lies on a circle of radius $2$, we represent its coordinates using the parameter $	heta$:
$$P = (2\cos	heta, 2\sin	heta)$$

For point $Q$ on the line $5x + y + 2 = 0$, if we let the $x$-coordinate be $\alpha$, the $y$-coordinate is determined by the line equation:
$$Q = (\alpha, -5\alpha - 2)$$

### The Two Pillars of the Bisector

A perpendicular bisector must satisfy two geometric conditions. First, the midpoint $M$ of segment $PQ$ must lie on the line $x - y + 1 = 0$. The midpoint is given by:
$$M = \left( \frac{2\cos	heta + \alpha}{2}, \frac{2\sin	heta - 5\alpha - 2}{2} ight)$$

Substituting these coordinates into the bisector equation $x - y + 1 = 0$ yields:
$$\cos	heta - \sin	heta + 3\alpha + 2 = 0$$

Second, the line $PQ$ must be perpendicular to the bisector $x - y + 1 = 0$. Since the slope of the bisector is $1$, the slope of $PQ$ must be $-1$:
$$\frac{(-5\alpha - 2) - 2\sin	heta}{\alpha - 2\cos	heta} = -1$$

Simplifying this slope condition leads to the second equation:
$$\sin	heta + \cos	heta + 2\alpha + 1 = 0$$

### The Algebraic Elimination

We now have a system of two linear equations in terms of $\alpha$:
1) $3\alpha = \sin	heta - \cos	heta - 2$
2) $2\alpha = -\sin	heta - \cos	heta - 1$

To eliminate $\alpha$, we multiply the first equation by $2$ and the second by $3$:
$$6\alpha = 2\sin	heta - 2\cos	heta - 4$$
$$6\alpha = -3\sin	heta - 3\cos	heta - 3$$

Equating the two expressions for $6\alpha$ results in:
$$2\sin	heta - 2\cos	heta - 4 = -3\sin	heta - 3\cos	heta - 3$$
$$\cos	heta + 5\sin	heta = 1$$

### The Trigonometric Climax

To solve $\cos	heta + 5\sin	heta = 1$, we use the half-angle substitution $t = 	an(\frac{	heta}{2})$. Applying the identities $\cos	heta = \frac{1-t^2}{1+t^2}$ and $\sin	heta = \frac{2t}{1+t^2}$:
$$\frac{1-t^2}{1+t^2} + 5\left(\frac{2t}{1+t^2}ight) = 1$$

Multiplying by $(1+t^2)$ gives:
$$1 - t^2 + 10t = 1 + t^2$$
$$2t^2 - 10t = 0$$

This yields two solutions for $t$: $t = 0$ and $t = 5$.

### Final Calculation

For $t = 0$, we have $\cos	heta = \frac{1-0}{1+0} = 1$. The corresponding abscissa of $P$ is $x_1 = 2(1) = 2$.

For $t = 5$, we have $\cos	heta = \frac{1-25}{1+25} = -\frac{24}{26} = -\frac{12}{13}$. The corresponding abscissa of $P$ is $x_2 = 2(-\frac{12}{13}) = -\frac{24}{13}$.

The sum of these abscissae is:
$$2 - \frac{24}{13} = \frac{26 - 24}{13} = \frac{2}{13}$$

The problem asks for $13$ times this sum:
$$13 	imes \frac{2}{13} = \mathbf{2}$$

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

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Parametric Form of Point $P$

General Form of Point $Q$

The Midpoint Condition

Simplifying the Midpoint Equation

The Perpendicularity Condition

Simplifying the Slope Equation

Eliminating $α$

The Trigonometric Equation

Half-Angle Substitution

Solving for $t$

Finding $cos θ$ Values

Abscissae of Point $P$

Sum of Abscissae

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Two Pillars of the Bisector

The Algebraic Elimination

The Trigonometric Climax

Final Calculation

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