Let the straight line y = 2x touch a circle with centre (0, α), α 0, and radius r at a point A₁. L

Q: Wait, why is the distance from the center to the line equal to the radius?

This is a fundamental property of circles! A tangent line is, by definition, perpendicular to the radius at the point of tangency. Therefore, the shortest distance from the center to the tangent line is exactly the radius $r$. It is the geometric bridge that connects the line's equation to the circle's properties.

Q: Why does the normal line have a slope of $-1/2$?

Great question! The tangent line is $y = 2x$, which has a slope of $m_1 = 2$. Since the normal line is perpendicular to the tangent, its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$. Thus, $2 \cdot m_2 = -1$, which gives $m_2 = -1/2$. It is all about the perpendicularity of the radius and the tangent.

Q: How do we know $C$ is the midpoint of $A_1B_1$?

The problem states that $A_1B_1$ is a diameter of the circle. By definition, a diameter is a chord that passes through the center of the circle. Since $A_1$ and $B_1$ are the endpoints of the diameter, the center $C$ must be the midpoint of the segment $A_1B_1$. It is a direct consequence of the definition of a diameter.

Question

Let the straight line $y = 2x$ touch a circle with centre $(0, \alpha)$, $\alpha > 0$, and radius $r$ at a point $A_{1}$. Let $B_{1}$ be the point on the circle such the line segment $A_{1}B_{1}$ is a diameter of the circle. Let $\alpha + r = 5 + \sqrt{5}$. Match each entry in List-I to the correct entries in List-II.

List-I:

P. $\alpha$ equals

Q. $r$ equals

R. $A_{1}$ equals

S. $B_{1}$ equals

List-II:

1. $(-2, 4)$

2. $\sqrt{5}$

3. $(-2, 6)$

4. $5$

5. $(2, 4)$

Sigma Percentile · Accepted Answer

### The Geometry of Tangency: A Journey into the Circle

Welcome, fellow explorer of mathematics. Today, we are going to peel back the layers of a beautiful coordinate geometry problem.

We are not just solving for variables; we are uncovering the hidden symmetry of a circle interacting with a line. Imagine you are standing at the center of a circle, $C(0, \alpha)$, looking out at a line $y = 2x$ that just barely kisses the edge of your circle at point $A_1$. This is the essence of tangency.

### Phase 1: The Tangency Condition

Our first task is to bridge the gap between the line and the circle. We know the line is $y = 2x$, or in standard form, $2x - y = 0$.

The center of our circle is $C(0, \alpha)$. The distance from a point $(x_0, y_0)$ to a line $ax + by + c = 0$ is given by the formula:

$$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$$

Applying this to our circle, the distance from $C(0, \alpha)$ to the line $2x - y = 0$ must be exactly the radius $r$. So, we have:

$$r = \frac{|2(0) - \alpha|}{\sqrt{2^2 + (-1)^2}}$$

Simplifying this, we get $r = \frac{|-\alpha|}{\sqrt{5}}$. Since the problem guarantees $\alpha > 0$, the absolute value simplifies beautifully to $r = \frac{\alpha}{\sqrt{5}}$, or $\alpha = r\sqrt{5}$. This is our first major breakthrough.

### Phase 2: Solving for the Circle

Now, we bring in the golden key provided by the problem: $\alpha + r = 5 + \sqrt{5}$. We have a system of two equations with two variables.

Substituting $\alpha = r\sqrt{5}$ into our relation, we get:

$$r\sqrt{5} + r = 5 + \sqrt{5}$$

Factoring out $r$ on the left, we have $r(\sqrt{5} + 1) = 5 + \sqrt{5}$. Look at the right side—it is a masterpiece of algebraic design. We can factor out $\sqrt{5}$ to get $\sqrt{5}(\sqrt{5} + 1)$.

The term $(\sqrt{5} + 1)$ cancels out perfectly from both sides, leaving us with the elegant result $r = \sqrt{5}$. Consequently, $\alpha = \sqrt{5} \cdot \sqrt{5} = 5$. Our circle is now fully defined: center $C(0, 5)$ and radius $r = \sqrt{5}$.

### Phase 3: Finding the Point of Tangency

With the circle defined, we seek the point $A_1$. We know the radius $CA_1$ is perpendicular to the tangent line $y = 2x$.

The slope of the tangent is $2$, so the slope of the normal line $CA_1$ must be $-1/2$. Since this normal line passes through $C(0, 5)$, its equation is:

$$y - 5 = -\frac{1}{2}(x - 0)$$

This simplifies to $x + 2y = 10$. To find $A_1$, we find the intersection of $y = 2x$ and $x + 2y = 10$.

Substituting $y = 2x$ into the normal equation, we get $x + 2(2x) = 10$, which means $5x = 10$, so $x = 2$. Then $y = 2(2) = 4$. Thus, **$A_1 = (2, 4)$**.

### Phase 4: The Diameter

Finally, we find $B_1$. We are told $A_1B_1$ is a diameter, meaning $C(0, 5)$ is the midpoint of $A_1(2, 4)$ and $B_1(x_B, y_B)$.

Using the midpoint formula, $\frac{x_B + 2}{2} = 0$ gives $x_B = -2$, and $\frac{y_B + 4}{2} = 5$ gives $y_B = 6$.

So, **$B_1 = (-2, 6)$**. We have successfully mapped the entire geometry of this problem. It is not just about the numbers; it is about the harmony of the circle and the line.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Setup

Point of Tangency

Distance to Tangent

Substituting Center

Simplifying Distance

Expressing $α$

Using Given Relation

Factoring the Equation

Solving for $r$ and $α$

Finding Point $A_{1}$

Equation of Normal

Intersection for $A_{1}$

Visualizing Diameter

Coordinates of $B_{1}$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Tangency

Phase 1

Phase 2

Phase 3

Phase 4

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