Let O be the centre of the circle x² + y² = r², where r √5/2. Suppose PQ is a chord of this circle

Q: Wait, why does the perpendicular bisector of chord PQ pass through the origin?

That is a fundamental property of circles! The perpendicular bisector of any chord in a circle always passes through the center of that circle. Since the origin O(0,0) is the center of our original circle, the line containing the perpendicular bisector must pass through the origin. This geometric insight is the key to unlocking the entire problem.

Q: Can we solve this using the distance of the chord from the center instead of the radical axis?

You absolutely could! You could calculate the distance from the origin to the line 2x + 4y = 5 and use the Pythagorean theorem to relate the radius r to the chord length. However, the radical axis method (S1 - S2 = 0) is much more elegant and powerful, especially in JEE Advanced problems where you might encounter more complex configurations.

Q: Why do we subtract the circle equations to get the common chord?

The equation S1 - S2 = 0 represents the locus of points where the power of the point with respect to both circles is equal. For two intersecting circles, this locus is exactly the line passing through their intersection points—which is the common chord. It is a beautiful algebraic shortcut that saves you from finding the intersection points P and Q explicitly.

Question

Let $O$ be the centre of the circle $x^2 + y^2 = r^2$, where $r > \frac{\sqrt{5}}{2}$. Suppose $PQ$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2x + 4y = 5$. If the centre of the circumcircle of the triangle $OPQ$ lies on the line $x + 2y = 4$, then the value of $r$ is ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a plane, looking at a circle centered at the origin, $O(0,0)$, with an unknown radius $r$. This is our primary world, $C_1$.

A chord $PQ$ slices through this circle, defined by the line $2x + 4y = 5$. A second world, the circumcircle $C_2$, wraps around the triangle formed by the origin and the endpoints of our chord.

### Locating the Heart of the Circumcircle

To understand the circumcircle $C_2$, we must first find its center. Geometry dictates that the center $C$ of a circumcircle lies at the intersection of the perpendicular bisectors of the triangle's sides.

For the triangle $	riangle OPQ$, the center $C$ must lie on the perpendicular bisector of the chord $PQ$. Because $PQ$ is a chord of the circle $C_1$, its perpendicular bisector must pass through the origin $O(0,0)$.

The slope of the line $PQ$ is $-\frac{1}{2}$, so the slope of the perpendicular bisector is $2$. Since it passes through the origin, the equation of this bisector is simply $y = 2x$.

We are given that the center $C$ also lies on the line $x + 2y = 4$. By solving the system of equations $y = 2x$ and $x + 2y = 4$, we find:

$$x + 2(2x) = 4 \Rightarrow 5x = 4 \Rightarrow x = \frac{4}{5}$$

Consequently, $y = \frac{8}{5}$. We have found the heart of our circumcircle: $C = (\frac{4}{5}, \frac{8}{5})$.

### The Radical Axis—The Elegant Shortcut

Since the circumcircle $C_2$ passes through the origin $O(0,0)$, the square of its radius $R^2$ is the squared distance from $O$ to $C$:

$$R^2 = \left(\frac{4}{5}ight)^2 + \left(\frac{8}{5}ight)^2 = \frac{16}{25} + \frac{64}{25} = \frac{80}{25} = \frac{16}{5}$$

The equation of this circumcircle $C_2$ is $(x - \frac{4}{5})^2 + (y - \frac{8}{5})^2 = \frac{16}{5}$, which expands to:

$$x^2 + y^2 - \frac{8}{5}x - \frac{16}{5}y = 0$$

We now have two circles: $S_1: x^2 + y^2 - r^2 = 0$ and $S_2: x^2 + y^2 - \frac{8}{5}x - \frac{16}{5}y = 0$. The line $PQ$ is the common chord of these two circles, found by $S_1 - S_2 = 0$:

$$(x^2 + y^2 - r^2) - (x^2 + y^2 - \frac{8}{5}x - \frac{16}{5}y) = 0 \Rightarrow \frac{8}{5}x + \frac{16}{5}y = r^2$$

### Final Calculation

We compare our derived equation $\frac{8}{5}x + \frac{16}{5}y = r^2$ with the given line $2x + 4y = 5$. Multiplying our derived equation by $\frac{5}{4}$ yields:

$$2x + 4y = \frac{5r^2}{4}$$

By comparing the constant terms, we see that $\frac{5r^2}{4} = 5$. This implies $r^2 = 4$.

Since the radius must be positive, the final result is **$r = 2$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Circumcircle $C_{2}$

Center of the Circumcircle

Equation of Perpendicular Bisector

Locating Center $C$

Solving for Center $C$

Radius of Circumcircle $C_{2}$

Equation of Circumcircle $C_{2}$

The Common Chord (Radical Axis)

Setting up $S_{1} - S_{2} = 0$

Finding the Value of $r$

The Sigma Insight: Director Circle and Family of Circles

Analyzing the Setup

Locating the Heart of the Circumcircle

The Radical Axis—The Elegant Shortcut

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Circumcircle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Center of the Circumcircle

Equation of Perpendicular Bisector

Locating Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Solving for Center .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Radius of Circumcircle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Equation of Circumcircle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​