If one of the diameters o of the circle x² + y² - 2x - 6y + 6 = 0 is a chord of another circle 'C' w

Q: Wait, why is the length of the chord equal to the diameter of the first circle? I thought it was just a part of it.

The problem explicitly states that 'one of the diameters' of the first circle is a chord of the second. This means the entire segment of length $2r_1$ is contained within the second circle. Therefore, the chord length is exactly $2r_1$.

Q: Why do we use the distance between centers as one of the legs of the triangle?

Great question! The line connecting the centers of two circles is always perpendicular to their common chord. This creates a right-angled triangle where the radius of the second circle is the hypotenuse, the distance between centers is one leg, and half the chord length is the other leg.

Q: Can I solve this using the equation of the chord instead of the distance formula?

You certainly could, but it would be much more tedious! Finding the equation of the diameter, then finding the perpendicular distance from $(2, 1)$ to that line, is mathematically equivalent to finding the distance between the two centers. The geometric approach is faster and less prone to calculation errors.

Q: Does the radius of the first circle change when it becomes a chord?

Not at all! The radius of the first circle is an intrinsic property of that circle. When we say it acts as a chord, we are simply using its length ($2r_1$) as a geometric constraint for the second circle.

Q: What if the centers were the same? Would the radius be different?

If the centers were the same, the distance $d$ would be zero. In that case, the chord would be a diameter of the second circle as well, and the radius $R$ would simply equal $r_1$. Our formula $R^2 = d^2 + r_1^2$ holds perfectly even in that degenerate case!

Question

If one of the diameters o of the circle $x^2 + y^2 - 2x - 6y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

Sigma Percentile · Accepted Answer

### Analyzing the First Circle

We begin with the equation $x^2 + y^2 - 2x - 6y + 6 = 0$. To understand this circle, we must bring it into the standard form $(x-h)^2 + (y-k)^2 = r^2$.

By completing the square for the $x$ and $y$ terms:
$$(x^2 - 2x + 1) + (y^2 - 6y + 9) = -6 + 1 + 9$$
$$(x - 1)^2 + (y - 3)^2 = 4$$

Comparing this to the standard form, we identify the center $C_1 = (1, 3)$ and the radius $r_1 = \sqrt{4} = 2$. We now know the first circle is centered at $(1, 3)$ with a radius of $2$.

### The Bridge Between Circles

The problem introduces a constraint: a diameter of this first circle acts as a chord for a second circle, $C_2$, centered at $(2, 1)$. The diameter of the first circle has a length of $2 	imes r_1 = 4$.

Because this entire diameter lies within the second circle, it serves as a chord of length $4$. The center of the first circle, $C_1(1, 3)$, is the midpoint of this diameter.

In any circle, the line segment from the center to the midpoint of a chord is perpendicular to that chord. Therefore, the distance $d$ between the center of the second circle $C_2(2, 1)$ and the center of the first circle $C_1(1, 3)$ represents the perpendicular distance from the center of the second circle to the chord.

### The Right Triangle of Destiny

To find the radius $R$ of the second circle, we construct a right-angled triangle. The hypotenuse is the radius $R$ of the second circle, connecting its center $(2, 1)$ to an endpoint of the chord.

One leg of this triangle is the distance $d$ between the two centers, and the other leg is half the length of the chord, which is equal to the radius $r_1 = 2$ of the first circle.

First, we calculate the distance $d$ between $C_1(1, 3)$ and $C_2(2, 1)$ using the distance formula:

$$d = \sqrt{(2 - 1)^2 + (1 - 3)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$

Now, we apply the **Pythagoras Theorem** to our triangle:

$$R^2 = d^2 + (r_1)^2$$

Substituting our known values:

$$R^2 = (\sqrt{5})^2 + (2)^2 = 5 + 4 = 9$$

### Final Calculation

Taking the square root of the result, we find the radius of the second circle:

**$R = 3$**

It is truly beautiful how the properties of circles interlock. By identifying the center and radius of the first circle, recognizing the geometric relationship between the centers and the chord, and applying the Pythagorean theorem, we have unraveled the mystery of the second circle.

Animated Solution for Mathematics - Circles: If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

Enter Numerical Value:

Visualized Solution

Equation of the First Circle

General Form of a Circle

Center of the First Circle

Radius of the First Circle

The Diameter as a Chord

Introducing the Second Circle

Distance Between Centers

Calculating Distance $d$

Forming a Right Triangle

Applying Pythagoras Theorem

Final Radius of Circle $C$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the First Circle

The Bridge Between Circles

The Right Triangle of Destiny

Final Calculation

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