If a circle passes through the point (a, b) and cuts the circle x² + y² = 4 orthogonally, then the l

Q: Wait, why do we use the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$ instead of $(x-h)^2 + (y-k)^2 = r^2$?

Great question! While the center-radius form is intuitive, the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ is much more powerful when dealing with orthogonality. The condition for orthogonality, $2g_1g_2 + 2f_1f_2 = c_1 + c_2$, is incredibly elegant and simplifies the algebra significantly compared to using the distance between centers and radii.

Q: Why does the constant $c$ turn out to be 4?

When we apply the orthogonality condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$ to our variable circle and the fixed circle $x^2 + y^2 - 4 = 0$, we identify $g_2 = 0$, $f_2 = 0$, and $c_2 = -4$. Substituting these into the formula gives $2g(0) + 2f(0) = c + (-4)$, which simplifies directly to $0 = c - 4$, or $c = 4$. It is a beautiful simplification that happens because the fixed circle is centered at the origin.

Q: How do I know the locus is a straight line?

Look at the final equation we derived: $2ax + 2by - (a^2 + b^2 + 4) = 0$. In this equation, $a$ and $b$ are constants (the coordinates of the fixed point). Since the equation is of the form $Ax + By + C = 0$, where $A=2a$, $B=2b$, and $C=-(a^2 + b^2 + 4)$, it is mathematically guaranteed to be a straight line!

Question

If a circle passes through the point $(a, b)$ and cuts the circle $x^2 + y^2 = 4$ orthogonally, then the locus of its centre is

Options:

(A) $2ax - 2by - (a^2 + b^2 + 4) = 0$

(B) $2ax + 2by - (a^2 + b^2 + 4) = 0$

(C) $2ax - 2by + (a^2 + b^2 + 4) = 0$

(D) $2ax + 2by + (a^2 + b^2 + 4) = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane. At the origin, there is a perfectly circular fountain, defined by the equation $x^2 + y^2 = 4$.

Nearby, there is a single, fixed point in space, marked by the coordinates $(a, b)$. Your mission is to track the center of a new, variable circle that must pass through that point $(a, b)$ and, crucially, must cut the fountain circle orthogonally.

We aim to find the path that the center of this new circle traces.

### The Anatomy of the Variable Circle

To begin, we define our variable circle in its general form:
$$x^2 + y^2 + 2gx + 2fy + c = 0$$

The center of this circle is located at $(-g, -f)$. We want to find the relationship between these coordinates as the circle changes.

Since the circle must pass through the fixed point $(a, b)$, we substitute these coordinates into our general equation:
$$a^2 + b^2 + 2ga + 2fb + c = 0$$

This equation acts as our anchor, linking the center parameters $(g, f)$ to the constant $c$ and the fixed point $(a, b)$.

### The Power of Orthogonality

Now, we introduce the condition of orthogonality. When two circles cut each other orthogonally, their radii at the point of intersection are perpendicular, leading to the algebraic condition:
$$2g_1g_2 + 2f_1f_2 = c_1 + c_2$$

For our fixed fountain circle, $x^2 + y^2 - 4 = 0$, we identify the parameters as $g_2 = 0$, $f_2 = 0$, and $c_2 = -4$.

Applying this to our variable circle, the terms involving $g$ and $f$ vanish:
$$2g(0) + 2f(0) = c + (-4)$$

This simplifies beautifully to $0 = c - 4$, or simply $c = 4$. The constant term of our variable circle is locked at $4$, regardless of where the circle is centered.

### Unveiling the Locus

With $c = 4$ determined, we return to our anchor equation:
$$a^2 + b^2 + 2ga + 2fb + 4 = 0$$

We define the center of our variable circle as $(x, y)$. Since the center is $(-g, -f)$, we substitute $g = -x$ and $f = -y$ into the equation:
$$a^2 + b^2 + 2(-x)a + 2(-y)b + 4 = 0$$

Rearranging the terms, we arrive at:
$$a^2 + b^2 - 2ax - 2by + 4 = 0$$

Finally, by shifting the terms to match the standard form, we obtain the equation of the locus:
$$2ax + 2by - (a^2 + b^2 + 4) = 0$$

This is a linear equation in $x$ and $y$. We have successfully proven that the locus of the center is a **straight line**.

Animated Solution for Mathematics - Circles: If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

Select Answer:

Visualized Solution

Visualizing the Fixed Elements

Defining the Variable Circle

Applying the Point Condition

The Orthogonality Tool

Parameters of the Fixed Circle

Computing the Constant $c$

Substituting $c$ Back

Setting Up the Locus Variables

Eliminating $g$ and $f$

The Final Locus Equation

The Sigma Insight: Condition for Orthogonality

Analyzing the Setup

The Anatomy of the Variable Circle

The Power of Orthogonality

Unveiling the Locus

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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 passes through the point (a,b) and cuts the circle x2+y2=4 orthogonally, then the locus of its centre is