If a circle touches y-axis at (0, 4) and passes through (2, 0) then which of the following can-not b

Q: Wait, why do we use the family of circles formula $(x-x_1)^2 + (y-y_1)^2 + \lambda L = 0$?

Great question! This formula is a powerful shortcut. It directly incorporates the tangency condition at a specific point. By using it, we avoid solving a system of three equations with three variables (the general form $x^2 + y^2 + 2gx + 2fy + c = 0$), making the algebra much cleaner and less prone to errors.

Q: How do I know the radius is 5?

Once you have the standard equation $x^2 + y^2 - 10x - 8y + 16 = 0$, you can complete the square to get $(x-5)^2 + (y-4)^2 = 25$. This shows the center is $(5, 4)$ and $R^2 = 25$, so $R = 5$. Alternatively, since the circle touches the y-axis ($x=0$), the radius must be the absolute value of the x-coordinate of the center, which is $|5| = 5$.

Q: What if the distance $d$ was less than $R$?

If the perpendicular distance $d$ from the center to the line is less than the radius $R$, the line is a secant, meaning it intersects the circle at two distinct points. If $d > R$, the line misses the circle entirely. Only when $d = R$ is the line perfectly tangent.

Question

If a circle touches y-axis at $(0, 4)$ and passes through $(2, 0)$ then which of the following can-not be the tangent to the circle

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You have a circle that is not just floating; it is delicately 'kissing' the y-axis at the point $(0, 4)$.

Because it touches the y-axis, the y-axis itself acts as a tangent line at that point. Furthermore, the circle swings down and passes through the point $(2, 0)$.

These two anchors—the point of tangency and the passing point—are all we need to define the circle's entire existence.

### The Magic of the Family of Circles

To find the equation of this circle, we do not need to struggle with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ immediately. Instead, we use the elegant 'family of circles' approach.

When a circle touches a line $L = 0$ at a point $(x_1, y_1)$, its equation is given by:

$$(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$$

Here, our line $L$ is the y-axis, which is $x = 0$, and our point of tangency is $(0, 4)$. Substituting these, we get:

$$(x - 0)^2 + (y - 4)^2 + \lambda x = 0$$

This equation is our skeleton key. It represents all circles touching the y-axis at $(0, 4)$.

### Unlocking the Parameter

We have one unknown, $\lambda$. But we have a second anchor: the point $(2, 0)$. Since the circle must pass through this point, it must satisfy our equation.

Substituting $x = 2$ and $y = 0$ into the equation, we get:

$$(2)^2 + (0 - 4)^2 + \lambda(2) = 0$$

This simplifies to $4 + 16 + 2\lambda = 0$, which means $2\lambda = -20$, or $\lambda = -10$.

With $\lambda$ found, our circle's DNA is complete:

$$x^2 + (y - 4)^2 - 10x = 0$$

Expanding this, we get:

$$x^2 + y^2 - 10x - 8y + 16 = 0$$

### The Heart of the Circle

From this general form, we can identify the center $(h, k)$ and the radius $R$. The center is $(5, 4)$, and the radius is:

$$R = \sqrt{5^2 + 4^2 - 16} = \sqrt{25 + 16 - 16} = 5$$

Now, we face the final challenge: determining which line is not a tangent. A line $Ax + By + C = 0$ is tangent to a circle if and only if the perpendicular distance $d$ from the center $(5, 4)$ to the line equals the radius $R = 5$.

The distance formula is:

$$d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$$

### The Tangency Test

Let us test the options:

**Option 1 ($4x - 3y + 17 = 0$):**

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

This is a tangent.

**Option 2 ($3x - 4y - 24 = 0$):**

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

This is also a tangent.

**Option 3 ($3x + 4y - 6 = 0$):**

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

This is another tangent.

**Option 4 ($4x + 3y - 8 = 0$):**

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

Since $4.8 
eq 5$, this line is **not a tangent**. We have successfully navigated the geometry and identified the outlier.

Animated Solution for Mathematics - Circles: If a circle touches y-axis at $(0, 4)$ and passes through $(2, 0)$ then which of the following can-not be the tangent to the circle

Select Answer:

Visualized Solution

Visualizing the Constraints

Family of Circles Equation

Finding the Parameter $λ$

Standard Equation of the Circle

Calculating the Radius

Tangency Condition: $d = R$

Testing Option 1

Testing Option 2

Testing Option 3

Testing Option 4 - The Answer

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Magic of the Family of Circles

Unlocking the Parameter

The Heart of the Circle

The Tangency Test

Similar Questions

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?

The centre of the circle passing through the point $(0, 1)$ and touching the curve $y = x^{2}$ at $(2, 4)$ is

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

Let $T_{1}, T_{2}$ be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle is:

A tangent is drawn to the parabola $y^{2} = 6 x$ which is perpendicular to the line $2 x + y = 1$ . Which of the following points does NOT lie on it?

The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

The centre of the circle passing through the point $(0, 1)$ and touching the parabola $y = x^{2}$ at the point $(2, 4)$ is:

The equations of the tangents drawn from the origin to the circle $x^{2} + y^{2} - 2 r x - 2 h y + h^{2} = 0$ , are

Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a circle touches y-axis at (0,4) and passes through (2,0) then which of the following can-not be the tangent to the circle