Let the abscissae of the two points P and Q be the roots of 2x² - rx + p = 0 and the ordinates of P

Q: Wait, the problem gives the equation for ordinates as $x^2 - sx - q = 0$. But ordinates are $y$-coordinates! Is this a typo?

That is a fantastic catch! In competitive exams, you will often see variables used as placeholders. The problem defines the ordinates as the roots of that specific quadratic. Whether the variable is written as $x$ or $y$ is irrelevant; the structure of the equation is what matters. Treat it as $y^2 - sy - q = 0$ and proceed with confidence.

Q: Why can't we just find the center and radius of the circle? That seems like the standard way to do it.

You absolutely could! You could find the midpoint of $PQ$ to get the center and the distance $PQ/2$ to get the radius. However, that path involves square roots and distance formulas, which are prone to calculation errors. The 'Diameter Form' is a shortcut that keeps the algebra clean and elegant. In JEE, choosing the path of least resistance is a skill in itself.

Q: How do I know when to compare the coefficients of the two equations?

Great question. You compare coefficients when you have two different expressions representing the same geometric object. We derived one equation from the roots, and the problem gave us another. Since they describe the same circle, they must be proportional. By matching them, we unlock the values of $r$, $s$, $p$, and $q$.

Question

Let the abscissae of the two points $P$ and $Q$ be the roots of $2x^2 - rx + p = 0$ and the ordinates of $P$ and $Q$ be the roots of $x^2 - sx - q = 0$. If the equation of the circle described on $PQ$ as diameter is $2(x^2 + y^2) - 11x - 14y - 22 = 0$, then $2r + s - 2q + p$ is equal to

Sigma Percentile · Accepted Answer

### The Dance of Algebra and Geometry

Welcome, future engineer. Today, we are going to unravel a problem that sits at the beautiful intersection of coordinate geometry and the theory of equations.

Often, students look at a problem like this—where roots of quadratics define the coordinates of a circle's diameter—and feel a sense of dread. But I want you to pause. Take a breath. This isn't a mess of variables; it is a symphony of structure. Let us walk through this together.

### Phase 1: The Roots as Coordinates

Imagine you are standing on the Cartesian plane. You have two points, $P$ and $Q$. We are told that the abscissae ($x$-coordinates) of these points are the roots of the quadratic equation $2x^2 - rx + p = 0$.

Let these roots be $x_1$ and $x_2$. Immediately, your mind should jump to Vieta's formulas. We know that for any quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product is $c/a$.

Applying this here, we get:

$$x_1 + x_2 = \frac{r}{2}$$

$$x_1x_2 = \frac{p}{2}$$

Now, look at the ordinates ($y$-coordinates). The problem gives us $y^2 - sy - q = 0$. Again, applying Vieta's, we find:

$$y_1 + y_2 = s$$

$$y_1y_2 = -q$$

We have now successfully translated the abstract algebra into concrete geometric properties of our points $P$ and $Q$.

### Phase 2: The Diameter Form

Here is where the magic happens. We need the equation of a circle with $PQ$ as its diameter. You could find the center, then the radius, then use $(x-h)^2 + (y-k)^2 = r^2$.

But why take the long road? We have a secret weapon: the **Diameter Form** of a circle. If $P(x_1, y_1)$ and $Q(x_2, y_2)$ are the endpoints of a diameter, the equation is simply:

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

Let us expand this. Multiplying the terms, we get:

$$x^2 - (x_1 + x_2)x + x_1x_2 + y^2 - (y_1 + y_2)y + y_1y_2 = 0$$

Look at that! It is practically begging us to substitute the sums and products we found in Phase 1. It is as if the problem was designed to fit these values perfectly.

### Phase 3: The Algebraic Bridge

Let us perform the substitution. Replacing the sums and products, our equation becomes:

$$x^2 - \left(\frac{r}{2}\right)x + \frac{p}{2} + y^2 - (s)y + (-q) = 0$$

To make this look like the standard form provided in the question, we multiply the entire equation by $2$. This gives us:

$$2x^2 - rx + p + 2y^2 - 2sy - 2q = 0$$

Rearranging this, we get:

$$2(x^2 + y^2) - rx - 2sy + p - 2q = 0$$

### Phase 4: The Final Comparison

We are at the finish line. The problem gives us the equation $2(x^2 + y^2) - 11x - 14y - 22 = 0$. Since this equation and our derived equation represent the same circle, their coefficients must match perfectly.

Comparing the $x$-coefficients, we see $-r = -11$, so $r = 11$. Comparing the $y$-coefficients, we see $-2s = -14$, so $s = 7$. Finally, the constant term: $p - 2q = -22$.

We are asked to find the value of $2r + s - 2q + p$. Let us group these:

$$2r + s + (p - 2q)$$

Substituting our values, we get:

$$2(11) + 7 + (-22) = 22 + 7 - 22$$

The $22$ and $-22$ cancel out beautifully, leaving us with the final answer of **7**.

See? When you break down the complexity into logical, manageable steps, the math doesn't just solve itself—it reveals its own internal harmony. Keep practicing this mindset, and you will conquer any problem JEE throws your way.

Enter Numerical Value:

Visualized Solution

Visualizing the Diameter

Abscissae as Roots

Sum and Product of Roots (x)

Ordinates as Roots

Sum and Product of Roots (y)

Diameter Form of a Circle

Expanding the Diameter Form

Substituting Sums and Products

Standardizing the Equation

Comparing with Given Equation

Equating Coefficients

Setting Up the Final Expression

Calculating the Final Value

The Sigma Insight: Standard and General Equation of a Circle

The Dance of Algebra and Geometry

Phase 1

Phase 2

Phase 3

Phase 4

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