Consider a circle C which touches the y-axis at (0, 6) and cuts off an intercept 6√5 on the x-axis.

Q: Wait, why is the center's $y$-coordinate 6?

The circle touches the $y$-axis at $(0, 6)$. Since the $y$-axis is a vertical line ($x=0$), the radius at the point of tangency must be horizontal. This means the center must share the same $y$-coordinate as the point of tangency, which is $6$.

Q: Why is the radius $r = |h|$?

The center is at $(h, 6)$. The distance from $(h, 6)$ to the $y$-axis ($x=0$) is the horizontal distance, which is $|h-0| = |h|$. Since the circle touches the $y$-axis, this distance is exactly the radius $r$.

Q: Why does the perpendicular from the center bisect the chord?

This is a fundamental theorem in circle geometry: the perpendicular from the center of a circle to any chord bisects the chord. It creates two congruent right-angled triangles, which is exactly what we need to solve the problem.

Question

Consider a circle $C$ which touches the $y$-axis at $(0, 6)$ and cuts off an intercept $6\sqrt{5}$ on the $x$-axis. Then the radius of the circle $C$ is equal to :

Options:

(A) $\sqrt{53}$

(B) 9

(C) 8

(D) $\sqrt{82}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden architecture of a circle.

Imagine standing on the Cartesian plane, looking at a circle that gracefully kisses the $y$-axis at the point $(0, 6)$. Because the circle touches the $y$-axis, the radius at that point must be horizontal.

This forces the center of our circle to lie on the horizontal line $y = 6$. Let us define the center as $C(h, 6)$.

### The Radius Connection

Now, consider the relationship between the center and the $y$-axis. Since the circle touches the $y$-axis, the horizontal distance from the center $(h, 6)$ to the $y$-axis ($x=0$) is exactly the radius $r$.

Thus, we find that $r = |h|$. This simple realization is the anchor for our entire calculation.

### The Chord and the Perpendicular

The problem states that the circle cuts the $x$-axis, creating an intercept of length $6\sqrt{5}$. Think of the $x$-axis as a chord of our circle.

We know from the fundamental theorems of geometry that a perpendicular dropped from the center of a circle to any chord bisects that chord. If we drop a perpendicular from our center $C(h, 6)$ to the $x$-axis, the length of this perpendicular is simply the $y$-coordinate of the center, which is $6$.

This perpendicular splits our chord of length $6\sqrt{5}$ into two equal segments, each of length $3\sqrt{5}$.

### The Pythagorean Bridge

Now, the beauty of the problem reveals itself. Connect the center $C(h, 6)$ to one of the endpoints of the $x$-intercept. We have just constructed a right-angled triangle!

The hypotenuse of this triangle is the radius $r$. The vertical leg is the perpendicular distance from the center to the $x$-axis, which is $6$. The horizontal leg is the half-chord length, $3\sqrt{5}$.

By the Pythagorean theorem, we have:

$$r^2 = 6^2 + (3\sqrt{5})^2$$

Let us calculate this with care: $6^2 = 36$, and $(3\sqrt{5})^2 = 9 	imes 5 = 45$.

Adding these together, we get:

$$r^2 = 36 + 45 = 81$$

Taking the positive square root, we find $r = 9$. We have arrived at our destination. The radius of the circle is **9**.

Animated Solution for Mathematics - Circles: Consider a circle $C$ which touches the $y$ -axis at $(0, 6)$ and cuts off an intercept $65$ on the $x$ -axis. Then the radius of the circle $C$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Tangency

Radius and Center Relationship

Analyzing the $x$ -intercept

Geometric Construction

Perpendicular Bisects the Chord

Forming the Right Triangle

Applying Pythagoras Theorem

Evaluating the Squares

Summing the Values

Finding the Final Radius

The Sigma Insight: Intercepts Made by a Circle

Analyzing the Setup

The Radius Connection

The Chord and the Perpendicular

The Pythagorean Bridge

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