If a circle passes through the point (a, b) and cuts the circle x² + y² = p² orthogonally, then the

Q: Wait, why is the locus a straight line? I expected a circle.

It is a common intuition to expect a curve, but remember that the locus is simply the set of all points satisfying a condition. In this case, the condition simplifies to a linear equation. Geometrically, this line is the radical axis of the system.

Q: Can I use the Pythagorean theorem instead of the orthogonality formula?

Absolutely! If you draw the triangle formed by the two centers and the point of intersection, the angle at the intersection is $90^\circ$. The square of the distance between centers equals the sum of the squares of the radii. Both methods lead to the same result.

Q: What if the point $(a, b)$ lies inside the fixed circle?

The problem assumes the existence of such a circle. If $(a, b)$ is inside, the circle would have to be imaginary or have a negative radius squared, which isn't physically possible in real Euclidean geometry. We assume the configuration is valid.

Question

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

Options:

(A) $x^2 + y^2 - 3ax - 4by + (a^2 + b^2 - p^2) = 0$

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

(C) $x^2 + y^2 - 2ax - 3by + (a^2 - b^2 - p^2) = 0$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every circle in the Cartesian plane can be described by the general equation:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

The center of this circle is located at $(-g, -f)$. Our goal is to track the path of this center as the circle varies.

Our variable circle must pass through the fixed point $(a, b)$. By substituting $x = a$ and $y = b$ into the general equation, we establish our first anchor:

$$a^2 + b^2 + 2ga + 2fb + c = 0 \quad 	ext{(Equation 1)}$$

### The Orthogonality Condition

Two circles cut orthogonally if the square of the distance between their centers equals the sum of the squares of their radii. Algebraically, for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$, the condition is:

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

Our fixed circle is $x^2 + y^2 - p^2 = 0$. Here, $g_2 = 0$, $f_2 = 0$, and $c_2 = -p^2$.

Substituting these values into the orthogonality condition, we obtain:

$$2g(0) + 2f(0) = c - p^2$$

This simplifies elegantly to $c = p^2$. The constant term of our variable circle is strictly determined by the radius of the fixed circle.

### The Locus Transformation

We now return to Equation 1 and substitute $c = p^2$:

$$a^2 + b^2 + 2ga + 2fb + p^2 = 0$$

We define the center of the variable circle as $(x, y)$. Since the center is $(-g, -f)$, we have the substitutions $g = -x$ and $f = -y$.

Substituting these into our equation yields:

$$a^2 + b^2 + 2(-x)a + 2(-y)b + p^2 = 0$$

Simplifying the expression, we arrive at the final equation for the locus:

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

This is a linear equation in terms of $x$ and $y$. Therefore, the locus of the center of the variable circle is a **straight line**.

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

Select Answer:

Visualized Solution

Visualizing the Setup

General Equation of Circle

Passing Through $(a, b)$

Orthogonality Condition

Applying Orthogonality

Finding $c$

Updating Equation 1

Locus of the Center

Final Substitution

Rearranging the Equation

The Sigma Insight: Condition for Orthogonality

Analyzing the Setup

The Orthogonality Condition

The Locus Transformation

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