A circle touches the y-axis at the point (0, 4) and passes through the point (2, 0). Which of the fo

Q: Wait, why is the y-coordinate of the center 4? Shouldn't it be related to the x-coordinate?

Great question! Think about the geometry of tangency. If a circle touches the y-axis at a specific point, the radius drawn to that point of tangency must be perpendicular to the y-axis. Since the y-axis is a vertical line, the radius at the point of tangency must be a horizontal line. This means the center of the circle must have the same y-coordinate as the point of tangency. Thus, if it touches at $(0, 4)$, the center must be at $y = 4$.

Q: Can I use the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ instead?

Absolutely! You can, but it is often more tedious. You would have three unknowns ($g, f, c$) and three conditions: the circle passes through $(0, 4)$, $(2, 0)$, and the tangency condition at $(0, 4)$. Using the center-radius form $(x-h)^2 + (y-k)^2 = R^2$ is much faster because it directly utilizes the geometric properties of the circle.

Q: How do I know if a line is a secant or a tangent?

The perpendicular distance $d$ from the center of the circle to the line is the key. If $d = R$, the line just 'kisses' the circle, making it a tangent. If $d R$, the line is too far away and doesn't touch the circle at all. In this problem, we are looking for the line that is NOT a tangent, so we are looking for the case where $d \neq R$.

Question

A circle touches the y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$. Which of the following lines is not a tangent to the circle?

Options:

(A) $4x - 3y + 17 = 0$

(B) $3x + 4y - 6 = 0$

(C) $4x + 3y - 8 = 0$

(D) $3x - 4y - 24 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of a Kissing Circle

Imagine you are standing on the Cartesian plane. You see a circle, a perfect, symmetric entity, gently kissing the y-axis at the point $(0, 4)$.

It is a beautiful moment of geometric contact. This circle then sweeps across the plane and passes through the point $(2, 0)$.

Our mission is to identify which of the given lines is not a tangent to this circle. To solve this, we must first unveil the circle's identity—its center and its radius.

### Phase 1: Unveiling the Center and Radius

Let us pause and visualize. If the circle touches the y-axis at $(0, 4)$, the radius at that point must be horizontal.

This forces the center of the circle to lie on the horizontal line $y = 4$. Let the center be $C(h, 4)$.

Because the circle touches the y-axis, the radius $R$ is simply the horizontal distance from the center to the y-axis, which is $|h|$. Since the circle is in the first quadrant (passing through $(2, 0)$), we can safely assume $h$ is positive, so $R = h$.

Our circle equation becomes:

$$ (x - h)^2 + (y - 4)^2 = h^2 $$

This is the soul of our circle, waiting to be defined.

### Phase 2: The Algebraic Anchor

We know the circle passes through $(2, 0)$. This is our anchor. By substituting $x = 2$ and $y = 0$ into our equation, we get:

$$ (2 - h)^2 + (0 - 4)^2 = h^2 $$

Let us expand this carefully:

$$ 4 - 4h + h^2 + 16 = h^2 $$

Notice the elegance here—the $h^2$ terms on both sides cancel out, leaving us with $20 - 4h = 0$. Solving this, we find $h = 5$.

Our circle is now fully revealed: the center is $C(5, 4)$ and the radius is $R = 5$. The equation is:

$$ (x - 5)^2 + (y - 4)^2 = 25 $$

### Phase 3: The Tangency Test

Now, we enter the final act. How do we determine if a line is a tangent? We use the perpendicular distance formula.

For a line $Ax + By + C = 0$, the distance $d$ from the center $(x_1, y_1)$ is given by:

$$ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} $$

If $d = R = 5$, the line is a tangent. If not, it is not. Let us test the options:

1. For $4x - 3y + 17 = 0$:

$$ d = \frac{|4(5) - 3(4) + 17|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 - 12 + 17|}{5} = \frac{25}{5} = 5 $$

This is a tangent.

2. For $3x + 4y - 6 = 0$:

$$ d = \frac{|3(5) + 4(4) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|15 + 16 - 6|}{5} = \frac{25}{5} = 5 $$

This is also a tangent.

3. For $4x + 3y - 8 = 0$:

$$ d = \frac{|4(5) + 3(4) - 8|}{\sqrt{4^2 + 3^2}} = \frac{|20 + 12 - 8|}{5} = \frac{24}{5} = 4.8 $$

Here, $d = 4.8$, which is strictly less than $5$. This line is a secant, not a tangent!

4. For $3x - 4y - 24 = 0$:

$$ d = \frac{|3(5) - 4(4) - 24|}{\sqrt{3^2 + (-4)^2}} = \frac{|15 - 16 - 24|}{5} = \frac{|-25|}{5} = 5 $$

This is a tangent.

### Conclusion

Through the power of coordinate geometry, we have systematically tested each line. The third option, **$4x + 3y - 8 = 0$**, fails the tangency condition because its distance from the center is $4.8$, not $5$.

It cuts through the circle, revealing itself as a secant. Always remember, in the world of JEE, the perpendicular distance formula is your most reliable scalpel for dissecting these problems. Keep practicing, and the intuition will become second nature!

Animated Solution for Mathematics - Circles: A circle touches the y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$ . Which of the following lines is not a tangent to the circle?

Select Answer:

Visualized Solution

Visualize the Given Conditions

Identify Center and Radius

Formulate the Circle Equation

Substitute the Given Point

Solve for the Radius $h$

Complete the Geometric Picture

The Tangency Condition

Test Option 1: $4 x - 3 y + 17 = 0$

Test Option 2: $3 x + 4 y - 6 = 0$

Test Option 3: $4 x + 3 y - 8 = 0$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of a Kissing Circle

Phase 1

Phase 2

Phase 3

Conclusion

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