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 – y + 1 = 0 is the perpendicular bisector of PQ, then 13 times the sum of abscissa of all such points P is

A perpendicular bisector of a line segment is a line that passes through the midpoint of the segment and is perpendicular to it. The fundamental property of a perpendicular bisector is that every point on it is equidistant from the endpoints of the segment. Mathematically, if line $L$ is the perpendicular bisector of segment $PQ$, and point $R$ lies on $L$, then $|RP| = |RQ|$. This property is extensively used in coordinate geometry to solve locus problems. The perpendicular bisector divides the plane into two half-planes, and it serves as the axis of symmetry for points $P$ and $Q$. In the context of reflection, if $L$ is the perpendicular bisector of $PQ$, then $P$ and $Q$ are reflections of each other about line $L$. The equation of perpendicular bisector can be found using the midpoint formula and the negative reciprocal of the slope. If a line segment has endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint is $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$ and the slope of $PQ$ is $m = \frac{y_2 – y_1}{x_2 – x_1}$. The perpendicular bisector passes through $M$ with slope $-\frac{1}{m}$. In many JEE problems, understanding that perpendicular bisector implies reflection symmetry simplifies the solution significantly. Applications include finding the circumcenter of a triangle (intersection of perpendicular bisectors of sides), determining the locus of points equidistant from two given points, and solving optimization problems in geometry. The concept extends to three dimensions where the perpendicular bisector becomes a plane equidistant from two points in space.

Reflection is a fundamental transformation in coordinate geometry that creates a mirror image of a geometric figure about a line called the axis of reflection. When a point $P(x_1, y_1)$ is reflected about a line $L: ax + by + c = 0$, the image point $P'(x_2, y_2)$ is found using the reflection formula. The line joining $P$ and $P’$ is perpendicular to $L$, and the midpoint of $PP’$ lies on $L$. The standard formula for reflection of point $(x_1, y_1)$ about line $ax + by + c = 0$ is: $\frac{x_2 – x_1}{a} = \frac{y_2 – y_1}{b} = -2\frac{ax_1 + by_1 + c}{a^2 + b^2}$. This formula is derived by using two conditions: perpendicularity and the fact that midpoint lies on the reflecting line. For special cases, reflection formulas are simpler. Reflection about x-axis $(y = 0)$: $(x, y) \to (x, -y)$. Reflection about y-axis $(x = 0)$: $(x, y) \to (-x, y)$. Reflection about origin: $(x, y) \to (-x, -y)$. Reflection about line $y = x$: $(x, y) \to (y, x)$. Reflection about line $y = -x$: $(x, y) \to (-y, -x)$. To reflect an entire line or curve, we can either reflect several points on it and find the new equation, or use parametric substitution method. When reflecting a line about another line, the angle that the original line makes with the axis of reflection equals the angle the reflected line makes (but on the opposite side). This is analogous to the law of reflection in optics. Reflection preserves distances and angles, making it an isometry. In JEE problems, reflection is often combined with circle-line intersection, distance optimization, and geometric construction problems.

A circle is the locus of all points equidistant from a fixed point called the center. The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x – h)^2 + (y – k)^2 = r^2$. When the center is at the origin, this simplifies to $x^2 + y^2 = r^2$. The general form of a circle equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, where the center is $(-g, -f)$ and radius is $\sqrt{g^2 + f^2 – c}$. For the circle to exist, we need $g^2 + f^2 – c > 0$. The parametric form of a circle $x^2 + y^2 = r^2$ is given by $x = r\cos\theta$ and $y = r\sin\theta$, where $\theta \in [0, 2\pi)$. This parametric representation is particularly useful when dealing with problems involving all points on a circle. Any point on the circle can be represented as $(r\cos\theta, r\sin\theta)$ for some angle $\theta$. Important properties of circles include: the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ represents a circle only if coefficient of $x^2$ equals coefficient of $y^2$ and there is no $xy$ term. The power of a point $(x_1, y_1)$ with respect to circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$. If power is positive, the point is outside; if zero, on the circle; if negative, inside. Circles are extensively used in JEE for problems involving tangents, normals, chord properties, family of circles, radical axis, and intersection with other conics. Understanding both algebraic and geometric properties is crucial for solving complex problems efficiently.

Finding the intersection points of a circle and a line is a fundamental problem in coordinate geometry. A straight line can intersect a circle at zero, one, or two points depending on the distance of the line from the center of the circle. Let the circle be $x^2 + y^2 = r^2$ (centered at origin) and the line be $ax + by + c = 0$. To find intersection points, we solve these equations simultaneously. One approach is to express one variable from the linear equation and substitute into the circle equation, resulting in a quadratic equation. For example, if line is $y = mx + d$, substituting into $x^2 + y^2 = r^2$ gives $x^2 + (mx + d)^2 = r^2$, which expands to $(1 + m^2)x^2 + 2mdx + (d^2 – r^2) = 0$. This is a quadratic in $x$. The discriminant $\Delta = 4m^2d^2 – 4(1 + m^2)(d^2 – r^2) = 4[(1 + m^2)r^2 – d^2(1 + m^2) + m^2d^2] = 4(1 + m^2)(r^2 – d^2)$ determines the number of intersection points. If $\Delta > 0$, two distinct real roots exist (line is a secant cutting circle at two points). If $\Delta = 0$, one repeated root exists (line is tangent to circle). If $\Delta < 0$, no real roots (line doesn't intersect circle). The perpendicular distance from center $(h, k)$ to line $ax + by + c = 0$ is $d = \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}}$. If $d < r$, line intersects at two points. If $d = r$, line is tangent. If $d > r$, no intersection. For sum and product of x-coordinates or y-coordinates of intersection points, Vieta’s formulas are extremely useful, as they provide these values without explicitly solving for the coordinates. This technique saves significant time in competitive exams like JEE.

Vieta’s formulas establish a relationship between the coefficients of a polynomial and the sum and product of its roots. For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, Vieta’s formulas state: sum of roots $\alpha + \beta = -\frac{b}{a}$ and product of roots $\alpha\beta = \frac{c}{a}$. These formulas are incredibly powerful in solving problems where we need the sum or product of roots without actually finding the individual roots. This is particularly useful in JEE where computational efficiency matters. For a cubic equation $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$, the formulas extend to: $\alpha + \beta + \gamma = -\frac{b}{a}$, $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$, and $\alpha\beta\gamma = -\frac{d}{a}$. In general, for polynomial $a_nx^n + a_{n-1}x^{n-1} + … + a_1x + a_0 = 0$ with roots $r_1, r_2, …, r_n$, Vieta’s formulas give: $\sum r_i = -\frac{a_{n-1}}{a_n}$, $\sum_{i