Find the equation of circle touching the line 2x + 3y + 1 = 0 at (1, -1) and cutting orthogonally th

Q: Wait, why do we use the family of circles formula? Can't we just find the center by drawing a perpendicular line from the point of tangency?

You absolutely can! You could find the slope of the tangent, find the slope of the normal, and then find the center by moving a distance $r$ along the normal. However, that approach is algebraically tedious and prone to errors. The family of circles formula is a powerful, elegant shortcut that directly incorporates the tangency condition, saving you time and mental energy for the rest of the problem.

Q: What does 'cutting orthogonally' actually mean in a geometric sense?

Geometrically, two circles are orthogonal if their tangents at the point of intersection are perpendicular to each other. This implies that the radii drawn to the point of intersection are also perpendicular. It's a beautiful property that simplifies the relationship between their centers and radii into the elegant algebraic condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.

Q: Why do we use the diameter form for the second circle? Is there another way?

The diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is the most direct way to write the equation of a circle when you know the endpoints of a diameter. You could find the midpoint (the center) and the distance between the points (the diameter), then use $(x - h)^2 + (y - k)^2 = r^2$, but the diameter form is faster and less likely to lead to calculation errors.

Question

Find the equation of circle touching the line $2x + 3y + 1 = 0$ at $(1, -1)$ and cutting orthogonally the circle having line segment joining $(0, 3)$ and $(-2, -1)$ as diameter.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane! Today, we are not just solving a problem; we are uncovering the hidden relationships between lines and circles.

Imagine you are standing on a vast, flat plane. You have a line, $L: 2x + 3y + 1 = 0$, stretching out before you. At the point $P(1, -1)$, a circle is perfectly balanced, touching this line.

This is a member of a vast, infinite family of circles, all kissing this line at exactly the same point $P$. To find the specific circle we need, we use the 'Family of Circles' equation: $(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$.

By substituting our point $P(1, -1)$ and the line $L$, we obtain:

$$(x - 1)^2 + (y + 1)^2 + \lambda(2x + 3y + 1) = 0$$

This is our starting point, the foundation upon which we will build our solution.

### The Second Circle: A Diameter Defined

Now, let us turn our attention to the second circle. We are given the endpoints of its diameter: $(0, 3)$ and $(-2, -1)$.

When you know the diameter, you hold the key to the circle's entire existence. We use the diameter form of the circle equation: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.

By plugging in our points, we get:

$$(x - 0)(x + 2) + (y - 3)(y + 1) = 0$$

Expanding this, we find the general form:

$$x^2 + y^2 + 2x - 2y - 3 = 0$$

Just like that, the mystery of the second circle is solved. We have its center and its radius, all hidden within these coefficients.

### The Orthogonality Condition: A Right-Angled Harmony

Here is where the magic happens. The problem states that our required circle cuts this second circle orthogonally.

This geometric harmony is captured by the powerful algebraic condition:

$$2g_1g_2 + 2f_1f_2 = c_1 + c_2$$

This is the bridge between our two circles. We extract the parameters $g, f,$ and $c$ from both equations.

For our first circle, expanding $(x - 1)^2 + (y + 1)^2 + \lambda(2x + 3y + 1) = 0$ gives:

$$x^2 + y^2 + (2\lambda - 2)x + (3\lambda + 2)y + (\lambda + 2) = 0$$

Thus, $g_1 = \lambda - 1$, $f_1 = \frac{3\lambda + 2}{2}$, and $c_1 = \lambda + 2$. For our second circle, $g_2 = 1$, $f_2 = -1$, and $c_2 = -3$.

Substituting these into our orthogonality condition, we get:

$$2(\lambda - 1)(1) + 2\left(\frac{3\lambda + 2}{2}\right)(-1) = (\lambda + 2) - 3$$

### The Final Synthesis

Now, we are in the home stretch. The equation simplifies as follows:

$$2\lambda - 2 - (3\lambda + 2) = \lambda - 1$$

This reduces to $-\lambda - 4 = \lambda - 1$, which yields $2\lambda = -3$, or $\lambda = -\frac{3}{2}$.

With $\lambda$ in hand, we return to our original family equation. Substituting $\lambda = -\frac{3}{2}$ back into our expanded general form:

$$x^2 + y^2 + \left(2\left(-\frac{3}{2}\right) - 2\right)x + \left(3\left(-\frac{3}{2}\right) + 2\right)y + \left(-\frac{3}{2} + 2\right) = 0$$

Simplifying this, we arrive at:

$$x^2 + y^2 - 5x - \frac{5}{2}y + \frac{1}{2} = 0$$

Multiplying by $2$ to clear the fractions, we reach our final, elegant destination:

**$$2x^2 + 2y^2 - 10x - 5y + 1 = 0$$**

Animated Solution for Mathematics - Circles: Find the equation of circle touching the line $2 x + 3 y + 1 = 0$ at $(1, - 1)$ and cutting orthogonally the circle having line segment joining $(0, 3)$ and $(- 2, - 1)$ as diameter.

Visualized Solution

Visualizing the Setup

The Family of Circles Concept

Substituting the Given Values

Expanding the General Equation

Rearranging into General Form

The Second Circle

Equation in Diameter Form

Expanding the Second Circle

The Orthogonality Condition

Extracting Parameters

Applying the Condition

Solving for Lambda

Substituting Lambda Back

The Final Equation

Summary and Key Takeaway

The Sigma Insight: Condition for Orthogonality

Analyzing the Setup

The Second Circle

The Orthogonality Condition

The Final Synthesis

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Visualized Solution

Visualizing the Setup

The Family of Circles Concept

Substituting the Given Values

Expanding the General Equation

Rearranging into General Form

The Second Circle

Equation in Diameter Form

Expanding the Second Circle

The Orthogonality Condition

Extracting Parameters

Applying the Condition

Solving for Lambda

Substituting Lambda Back

The Final Equation

Summary and Key Takeaway

The Sigma Insight: Condition for Orthogonality

Analyzing the Setup

The Second Circle

The Orthogonality Condition

The Final Synthesis

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