A circle S passes through the point (0, 1) and is orthogonal to the circles (x - 1)² + y² = 16 and x

Q: Wait, why do we use $2g_1g_2 + 2f_1f_2 = c_1 + c_2$? Can't we just use the distance between centers?

You *could* use the Pythagorean theorem on the triangle formed by the two centers and the intersection point, but that's the long way around! The orthogonality condition is derived directly from that geometry. Using the condition is faster and less prone to calculation errors.

Q: Wait, why is the center $(-g, -f)$ and not $(g, f)$?

It's a matter of convention in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$. If you complete the square, you get $(x+g)^2 + (y+f)^2 = g^2 + f^2 - c$. The center is clearly $(-g, -f)$. Don't let the signs trip you up!

Q: Do I really need to expand $(x - 1)^2 + y^2 = 16$?

Absolutely. The orthogonality condition requires the coefficients $g_2, f_2, c_2$ from the standard general form. You cannot read them directly from the $(x-h)^2 + (y-k)^2 = r^2$ form. Expansion is the bridge to the solution.

Question

A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x - 1)^2 + y^2 = 16$ and $x^2 + y^2 = 1$. Then

Options:

(A) radius of $S$ is 8

(B) radius of $S$ is 7

(C) centre of $S$ is $(-7, 1)$

(D) centre of $S$ is $(-8, 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the circle $S$ that intersects two given circles at right angles, we utilize the general equation of a circle:

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

Our objective is to determine the specific values of the parameters $g$, $f$, and $c$.

### The First Constraint: A Point of Passage

The circle $S$ passes through the point $(0, 1)$. By substituting $x = 0$ and $y = 1$ into the general equation, we obtain:

$$0^2 + 1^2 + 2g(0) + 2f(1) + c = 0$$

This simplifies to the following anchor equation:

$$1 + 2f + c = 0 \quad 	ext{(i)}$$

### The Power of Orthogonality

Two circles are orthogonal if they satisfy the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. For the first circle, $x^2 + y^2 = 1$, the center is $(0, 0)$ and the constant term is $c_2 = -1$.

Substituting these values into the orthogonality condition:

$$2g(0) + 2f(0) = c + (-1)$$

This simplifies directly to:

$$c = 1$$

### The Second Encounter

We now consider the second circle, $(x - 1)^2 + y^2 = 16$. Expanding this equation yields:

$$x^2 - 2x + 1 + y^2 = 16 \Rightarrow x^2 + y^2 - 2x - 15 = 0$$

Here, the parameters are $g_2 = -1$, $f_2 = 0$, and $c_2 = -15$. Applying the orthogonality condition again:

$$2g(-1) + 2f(0) = c - 15$$

Substituting $c = 1$ into this expression:

$$-2g = 1 - 15 \Rightarrow -2g = -14 \Rightarrow g = 7$$

### The Final Piece

With $c = 1$ and $g = 7$, we return to Equation (i):

$$1 + 2f + 1 = 0 \Rightarrow 2f = -2 \Rightarrow f = -1$$

The equation of the circle $S$ is therefore:

$$x^2 + y^2 + 14x - 2y + 1 = 0$$

The center of the circle is $(-g, -f)$, which is $(-7, 1)$. The radius $r$ is calculated as:

$$r = \sqrt{g^2 + f^2 - c} = \sqrt{7^2 + (-1)^2 - 1} = \sqrt{49 + 1 - 1} = \sqrt{49} = 7$$

We have successfully determined the circle's properties, confirming the **center is $(-7, 1)$** and the **radius is $7$**.

Animated Solution for Mathematics - Circles: A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x - 1)^{2} + y^{2} = 16$ and $x^{2} + y^{2} = 1$ . Then

Select Answer:

Visualized Solution

$S : x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

Point $(0, 1)$ Constraint

Substituting $(0, 1)$

Equation $1$

Orthogonality Condition

Orthogonality with $x^{2} + y^{2} = 1$

Solving for $c$

Second Orthogonal Circle

Expanding $(x - 1)^{2} + y^{2} = 16$

Applying Condition Again

Solving for $g$

Solving for $f$

Centre of Circle $S$

Radius of Circle $S$

Final Answer

The Sigma Insight: Condition for Orthogonality

Analyzing the Setup

The First Constraint

The Power of Orthogonality

The Second Encounter

The Final Piece

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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; } A circle S passes through the point (0,1) and is orthogonal to the circles (x−1)2+y2=16 and x2+y2=1. Then

Select Answer:

Visualized Solution

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