Let a circle C of radius 5 lie below the x-axis. The line L1 = 4x + 3y - 2 = 0 passes through the ce

Q: Wait, why do we assume $L_1$ is the normal to the circle at $Q$?

That is a fantastic observation! In geometry, a fundamental theorem states that the radius of a circle is always perpendicular to the tangent at the point of contact. Since $L_1$ passes through the center $P$ and is perpendicular to the tangent $L_2$ at $Q$, it must contain the radius $PQ$. Therefore, $L_1$ is indeed the normal line at $Q$.

Q: How do I know which of the two candidate centers to choose?

The problem provides a crucial constraint: the circle lies entirely below the x-axis. This means the highest point of the circle, which is at $y = y_P + r$, must be less than zero. By testing both candidates, you will find that one results in a circle that crosses the x-axis, while the other stays safely below it. Always look for these 'hidden' constraints in coordinate geometry problems!

Q: Can I use the distance formula directly without finding the center $P$?

Unfortunately, no. The distance we need is from the center $P$ to the line $L_3$. Without the specific coordinates of $P$, we cannot apply the distance formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. Finding $P$ is the essential bridge to the final answer.

Question

Let a circle $C$ of radius 5 lie below the x-axis. The line $L_1 = 4x + 3y - 2 = 0$ passes through the centre $P$ of the circle $C$ and intersects the line $L_2 : 3x - 4y - 11 = 0$ at $Q$. The line $L_2$ touches $C$ at the point $Q$. Then the distance of $P$ from the line $5x - 12y + 51 = 0$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a circle $C$ with a radius of $r=5$ positioned entirely below the $x$-axis. We are provided with two lines:
$L_1: 4x + 3y + 2 = 0$
$L_2: 3x - 4y - 11 = 0$

Our objective is to determine the perpendicular distance from the center of this circle to a third line, $L_3: 5x - 12y + 51 = 0$.

### The Orthogonal Dance

First, we examine the slopes of $L_1$ and $L_2$. The slope of $L_1$ is $m_1 = -\frac{4}{3}$, and the slope of $L_2$ is $m_2 = \frac{3}{4}$.

Since the product of the slopes is:
$$m_1 	imes m_2 = \left(-\frac{4}{3}ight) 	imes \left(\frac{3}{4}ight) = -1$$

This confirms that $L_1$ and $L_2$ are perpendicular. Because $L_1$ passes through the center $P$ and is perpendicular to the tangent $L_2$ at the point of contact $Q$, $L_1$ acts as the normal to the circle at $Q$.

### Finding the Point of Contact

We solve the system of equations for $L_1$ and $L_2$ to find their intersection point $Q$:
$$4x + 3y + 2 = 0$$
$$3x - 4y - 11 = 0$$

Solving this system yields the point of contact $Q(1, -2)$. This point anchors the circle to the line $L_2$.

### The Parametric Leap

The center $P$ lies on $L_1$ at a distance of $r=5$ from $Q$. Given the slope of $L_1$ is $-\frac{4}{3}$, we have $	an 	heta = -\frac{4}{3}$, which implies $\cos 	heta = \pm \frac{3}{5}$ and $\sin 	heta = \mp \frac{4}{5}$.

Using the parametric coordinates $x = x_Q \pm r \cos 	heta$ and $y = y_Q \pm r \sin 	heta$, we identify two potential centers:
$$P_1 = (1 - 3, -2 + 4) = (-2, 2)$$
$$P_2 = (1 + 3, -2 - 4) = (4, -6)$$

### The Constraint Filter

The problem states the circle lies entirely below the $x$-axis. This requires the highest point of the circle, defined by $y_P + r$, to be less than zero.

For $P_1(-2, 2)$:
$$y_P + r = 2 + 5 = 7 > 0$$

For $P_2(4, -6)$:
$$y_P + r = -6 + 5 = -1 < 0$$

Thus, the valid center of the circle is **$P(4, -6)$**.

### Final Calculation

We calculate the perpendicular distance $d$ from $P(4, -6)$ to the line $L_3: 5x - 12y + 51 = 0$ using the formula:
$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting the coordinates of $P$ and the coefficients of $L_3$:
$$d = \frac{|5(4) - 12(-6) + 51|}{\sqrt{5^2 + (-12)^2}}$$

$$d = \frac{|20 + 72 + 51|}{\sqrt{25 + 144}} = \frac{143}{\sqrt{169}} = \frac{143}{13}$$

The final distance is **$11$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Establishing Orthogonality

Finding Intersection Point $Q$

Locating the Center $P$

Applying the 'Below x-axis' Constraint

The Final Target: Distance to $L_{3}$

Distance Formula Setup

Atomic Computation

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Orthogonal Dance

Finding the Point of Contact

The Parametric Leap

The Constraint Filter

Final Calculation

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