The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through t

Q: Wait, why is the y-coordinate of the center equal to 2? I thought it could be anything.

Great question! Think about the tangency condition. If a circle touches the y-axis at $(0, 2)$, the radius at that point must be perpendicular to the y-axis. A line perpendicular to the vertical y-axis is a horizontal line. Since the radius connects the center to $(0, 2)$, that horizontal line must pass through the center. Therefore, the center must share the same y-coordinate as the point of tangency, which is $y = 2$.

Q: Why did we set $r^2 = h^2$? Is the radius always equal to the x-coordinate of the center?

Not exactly the x-coordinate, but the absolute value of it! The radius is the perpendicular distance from the center $(h, 2)$ to the y-axis (the line $x = 0$). The distance from a point $(h, k)$ to the y-axis is simply $|h|$. So, $r = |h|$, which means $r^2 = h^2$. This is a powerful simplification that turns a two-variable problem into a one-variable problem.

Q: Why can't we just use the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$?

You absolutely can! However, the center-radius form $(x - h)^2 + (y - k)^2 = r^2$ is often more intuitive when you are given geometric constraints like tangency. Using the general form would require you to solve for three unknowns ($g, f, c$) using three conditions, which is much more algebraic heavy lifting. Always choose the tool that aligns with the geometric information provided.

Question

The circle passing through the point $(-1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

Options:

(A) $(-\frac{3}{2}, 0)$

(B) $(-\frac{5}{2}, 2)$

(C) $(-\frac{3}{2}, \frac{5}{2})$

(D) $(-4, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of mathematics. Today, we are not just solving a problem; we are reconstructing a geometric object from the clues it leaves behind.

We are given a circle that touches the y-axis at $(0, 2)$ and passes through the point $(-1, 0)$. This is a classic JEE Advanced scenario where the beauty lies in translating geometric constraints into algebraic elegance.

Imagine a circle resting gently against the y-axis at the height of $y = 2$. Because the tangent is vertical, the radius drawn to this point must be perfectly horizontal.

This implies that the center of our circle must lie on the horizontal line $y = 2$. If we denote the center as $C(h, 2)$, we have already locked in the y-coordinate.

The radius is the perpendicular distance from the center to the y-axis. Since the center is at a horizontal distance of $h$ from the y-axis, the radius $r$ must be equal to $|h|$.

This gives us the relationship:
$$r^2 = h^2$$

### Constructing the Algebraic Bridge

With our center $C(h, 2)$ and radius $r^2 = h^2$ defined, we can now construct the standard equation of the circle using the form $(x - h)^2 + (y - k)^2 = r^2$.

Substituting our known values, we obtain:
$$(x - h)^2 + (y - 2)^2 = h^2$$

We now use our second clue: the circle passes through the point $A(-1, 0)$. If this point lies on the circle, it must satisfy our equation.

Substituting $x = -1$ and $y = 0$ into the equation:
$$(-1 - h)^2 + (0 - 2)^2 = h^2$$

Expanding the terms, we note that $(-1 - h)^2 = (1 + h)^2 = 1 + 2h + h^2$. The second term, $(0 - 2)^2$, is simply $4$.

Thus, our equation becomes:
$$1 + 2h + h^2 + 4 = h^2$$

### The Elegance of Cancellation

Observe the equation $1 + 2h + h^2 + 4 = h^2$. The $h^2$ terms appear on both sides and cancel out completely.

This leaves us with a simple linear equation:
$$2h + 5 = 0 \Rightarrow h = -\frac{5}{2}$$

We have found our center: $C(-\frac{5}{2}, 2)$. Substituting this back into our circle equation, we get:
$$(x + \frac{5}{2})^2 + (y - 2)^2 = (-\frac{5}{2})^2$$

Squaring the right side gives $\frac{25}{4}$. Expanding into the general form:
$$x^2 + 5x + \frac{25}{4} + y^2 - 4y + 4 = \frac{25}{4}$$

The $\frac{25}{4}$ terms cancel out, leaving us with the elegant final equation:
$$x^2 + y^2 + 5x - 4y + 4 = 0$$

### The Final Verification

To verify if a point lies on this circle, we test the point $(-4, 0)$ by substituting $x = -4$ and $y = 0$ into our final equation:
$$(-4)^2 + (0)^2 + 5(-4) - 4(0) + 4$$

Calculating this, we get:
$$16 + 0 - 20 - 0 + 4 = 0$$

The equation is satisfied. We have successfully navigated the geometry, the algebra, and the verification to confirm the circle's properties.

Animated Solution for Mathematics - Circles: The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

Select Answer:

Visualized Solution

Visualizing the Given Points

The Tangency Condition

Defining Center and Radius

Standard Equation of the Circle

Substituting Point $A (- 1, 0)$

Expanding the Equation

Solving for $h$

The Final Equation

Expanding to General Form

Verifying the Options

Final Verification

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Constructing the Algebraic Bridge

The Elegance of Cancellation

The Final Verification

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The circle passing through the point (−1,0) and touching the y-axis at (0,2) also passes through the point