Let a circle C touch the lines L1 : 4x - 3y + K1 = 0 and L2 : 4x - 3y + K2 = 0, K1, K2 ∈ R. If a lin

Q: Wait, why is the center of the circle the midpoint of the segment PQ?

That is a fantastic question! Because the circle is tangent to both parallel lines, the line segment connecting the points of tangency is a diameter of the circle. Any line passing through the center that intersects both parallel lines will have its segment between those lines bisected by the center of the circle due to the symmetry of the circle relative to the parallel tangents.

Q: Can I just use the distance from the center to one of the lines to find the radius?

Absolutely! Once you find the center $(h, k)$ and the equation of one of the lines, the perpendicular distance from the center to that line is indeed the radius $r$. It is a great way to double-check your work, though finding the distance between the two parallel lines first is often faster.

Q: What if I made a mistake with the signs when calculating K1 and K2?

It happens to the best of us! Always remember that the equation is $4x - 3y + K = 0$. When you substitute $x = -1$ and $y = 2$, you get $4(-1) - 3(2) + K_1 = 0$, which is $-4 - 6 + K_1 = 0$. Be very careful with the negative signs—they are the most common place for errors in coordinate geometry.

Question

Let a circle $C$ touch the lines $L_1 : 4x - 3y + K_1 = 0$ and $L_2 : 4x - 3y + K_2 = 0, K_1, K_2 \in \mathbb{R}$. If a line passing through the centre of the circle $C$ intersects $L_1$ at $(-1, 2)$ and $L_2$ at $(3, -6)$, then the equation of the circle $C$ is

Options:

(A) $(x - 1)^2 + (y - 2)^2 = 4$

(B) $(x + 1)^2 + (y - 2)^2 = 4$

(C) $(x - 1)^2 + (y + 2)^2 = 16$

(D) $(x - 1)^2 + (y - 2)^2 = 16$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two parallel lines, $L_1: 4x - 3y + K_1 = 0$ and $L_2: 4x - 3y + K_2 = 0$. A circle $C$ is perfectly wedged between them, touching both.

A transversal line passes through the center of this circle, intersecting $L_1$ at $P(-1, 2)$ and $L_2$ at $Q(3, -6)$. Our mission is to find the equation of this circle.

### Unlocking the Constants

Before we can find the circle, we need to determine the values of $K_1$ and $K_2$. Since point $P(-1, 2)$ lies on $L_1$, it must satisfy the equation $4x - 3y + K_1 = 0$.

Substituting the coordinates, we get:
$$4(-1) - 3(2) + K_1 = 0 \Rightarrow -4 - 6 + K_1 = 0 \Rightarrow K_1 = 10$$

Similarly, for point $Q(3, -6)$ on $L_2$, we have $4(3) - 3(-6) + K_2 = 0$. This simplifies to:
$$12 + 18 + K_2 = 0 \Rightarrow K_2 = -30$$

We have successfully pinned down our lines: $4x - 3y + 10 = 0$ and $4x - 3y - 30 = 0$.

### The Diameter of the Circle

The perpendicular distance between two parallel lines $ax + by + K_1 = 0$ and $ax + by + K_2 = 0$ is given by the formula:
$$d = \frac{|K_1 - K_2|}{\sqrt{a^2 + b^2}}$$

In our case, $a = 4$ and $b = -3$. The distance between the lines is the diameter of the circle, $2r$.

Calculating the diameter:
$$2r = \frac{|10 - (-30)|}{\sqrt{4^2 + (-3)^2}} = \frac{|40|}{\sqrt{16 + 9}} = \frac{40}{5} = 8$$

If the diameter is $8$, then the radius $r$ is $4$, and $r^2 = 16$.

### Locating the Center

The center of the circle lies on the line passing through $P$ and $Q$. Because the circle is tangent to both parallel lines, its center must be exactly halfway between them.

By the symmetry of the circle, the center $M(h, k)$ is the midpoint of the segment $PQ$. Using the midpoint formula:
$$h = \frac{-1 + 3}{2} = 1, \quad k = \frac{2 + (-6)}{2} = -2$$

The center of the circle is at $(1, -2)$.

### Final Assembly

We have the center $(1, -2)$ and the radius squared $r^2 = 16$. The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.

Substituting our values, we get:
$$(x - 1)^2 + (y - (-2))^2 = 16$$

Simplifying this, we arrive at the final equation:
$$(x - 1)^2 + (y + 2)^2 = 16$$

This is the equation of our circle. It is elegant, precise, and perfectly matches our geometric intuition.

Select Answer:

Visualized Solution

Visualizing the Setup

Finding the Constants $K_{1}$ and $K_{2}$

Substituting $P$ into $L_{1}$

Calculating $K_{1}$

Substituting $Q$ into $L_{2}$

Calculating $K_{2}$

Diameter from Parallel Tangents

Setting up the Distance Formula

Calculating the Radius

Locating the Center of the Circle

Applying the Midpoint Formula

Calculating Center Coordinates

Setting up the Circle Equation

The Final Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Unlocking the Constants

The Diameter of the Circle

Locating the Center

Final Assembly

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