Let O be the origin and OP and OQ be the tangents to the circle x² + y² - 6x + 4y + 8 = 0 at the poi

Q: Why can't we just find the coordinates of $P$ and $Q$ and then find the circumcircle?

While theoretically possible, finding the intersection of the tangents with the circle involves solving a system of quadratic equations, which is algebraically tedious and prone to error. The JEE Advanced exam rewards those who spot the geometric shortcut: the property that $O, P, C, Q$ are concyclic.

Q: Wait, why is $OC$ the diameter? I thought the diameter was just a line through the center.

A diameter is a chord passing through the center. In this case, because $\angle OPC = 90^\circ$, the circle passing through $O, P,$ and $C$ must have $OC$ as its diameter by the converse of Thales' Theorem (an angle subtended by a diameter at the circumference is $90^\circ$).

Q: Is there only one circumcircle for $\triangle OPQ$?

Yes, for any three non-collinear points, there is exactly one unique circle that passes through them. By proving that $O, P, Q,$ and $C$ are concyclic, we simplify the problem of finding the circumcircle of $\triangle OPQ$ to finding the circle with diameter $OC$.

Question

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2 + y^2 - 6x + 4y + 8 = 0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$, then a value of $\alpha$ is

Options:

(A) $\frac{3}{2}$

(B) $-\frac{1}{2}$

(C) $\frac{5}{2}$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the origin $O(0, 0)$, looking out at a circle defined by the equation $x^2 + y^2 - 6x + 4y + 8 = 0$. You draw two lines from the origin that just graze the edge of this circle, touching it at points $P$ and $Q$.

You are asked to find the circumcircle of the triangle $	riangle OPQ$. At first glance, this seems like a nightmare of coordinate geometry—calculating the intersection points $P$ and $Q$ seems like a trap designed to waste your time.

But wait! Let us pause and look for the hidden symmetry.

### Unmasking the Circle

First, we must understand our circle. By comparing $x^2 + y^2 - 6x + 4y + 8 = 0$ with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we find $g = -3$ and $f = 2$.

The center $C$ is at $(-g, -f)$, which gives us $C(3, -2)$. This is our anchor point.

### The Geometric Revelation

Here is the secret that separates the top rankers from the rest. The radius $CP$ is perpendicular to the tangent $OP$.

This means $\angle OPC = 90^{\circ}$. Similarly, $\angle OQC = 90^{\circ}$.

If you look at the quadrilateral $OPCQ$, you see two opposite angles that are $90^{\circ}$. More importantly, the points $O, P, C,$ and $Q$ all lie on a circle where $OC$ is the diameter.

This is because the segment $OC$ subtends a right angle at both $P$ and $Q$. This is the 'Aha!' moment; we do not need $P$ or $Q$, we only need $O$ and $C$.

### The Algebraic Dance

With $O(0, 0)$ and $C(3, -2)$ as the endpoints of our diameter, the equation of the circumcircle is given by the diameter form:
$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

Substituting our coordinates, we get:
$$(x - 0)(x - 3) + (y - 0)(y + 2) = 0$$

This simplifies beautifully to:
$$x^2 + y^2 - 3x + 2y = 0$$

### The Final Resolution

The problem states that this circle passes through the point $(\alpha, \frac{1}{2})$. If a point lies on a curve, it must satisfy the equation.

So, we substitute $x = \alpha$ and $y = \frac{1}{2}$:
$$\alpha^2 + \left( \frac{1}{2} ight)^2 - 3\alpha + 2\left( \frac{1}{2} ight) = 0$$

Simplifying this, we get:
$$\alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0 \Rightarrow \alpha^2 - 3\alpha + \frac{5}{4} = 0$$

Multiplying by $4$ to clear the fraction, we arrive at:
$$4\alpha^2 - 12\alpha + 5 = 0$$

Factoring this quadratic, we find:
$$(2\alpha - 5)(2\alpha - 1) = 0$$

Thus, $\alpha = \frac{5}{2}$ or $\alpha = \frac{1}{2}$. Looking at our options, **$\alpha = \frac{5}{2}$** is the clear winner. You have just navigated a complex geometry problem by using the elegance of circle properties rather than brute-force calculation. That is the JEE way.

Animated Solution for Mathematics - Circles: Let $O$ be the origin and $O P$ and $O Q$ be the tangents to the circle $x^{2} + y^{2} - 6 x + 4 y + 8 = 0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $O P Q$ passes through the point $(α, \frac{1}{2})$ , then a value of $α$ is

Select Answer:

Visualized Solution

Equation of the Given Circle

Center of the Circle

Tangents $O P$ and $O Q$

Circumcircle of $△ O P Q$

Diameter $O C$

Diameter Form Equation

Simplifying the Equation

Point $(α, \frac{1}{2})$ on the Circle

Forming the Equation in $α$

Standard Quadratic Form

Solving for $α$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Unmasking the Circle

The Geometric Revelation

The Algebraic Dance

The Final Resolution

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Comprehension Passage

The equation of circle $C$ is

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Let P be a point on the parabola, $x^{2} = 4 y$ . If the distance of P from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at P, is :

Select Answer:

Visualized Solution

Equation of the Given Circle

Center of the Circle

Tangents .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OP and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OQ

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

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