Question

Let \(C\) be a circle having centre in the first quadrant and touching the \(x\)-axis at a distance of \(3\) units from the origin. If the circle \(C\) has an intercept of length \(6\sqrt{3}\) on \(y\)-axis, then the length of the chord of the circle \(C\) on the line \(x-y=3\) is:

• A. \(8\)
• B. \(6\)
• C. \(6\sqrt{2}\)
• D. \(8\sqrt{2}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Question
We are given information about a circle \(C\) and need to find the length of a chord it makes with a specific line. We first need to determine the circle's equation (its center and radius) from the given geometric properties.
Step 2: Key Formula or Approach
1. If a circle touches the x-axis at a point \((h,0)\), its center must be at \((h,k)\) and its radius will be \(|k|\). Since the center is in the first quadrant, the center is \((h,r)\) and radius is \(r\).
2. The length of an intercept made by a circle with radius \(r\) on a line is \(2\sqrt{r^2 - d^2}\), where \(d\) is the perpendicular distance from the center of the circle to the line.
3. The perpendicular distance from a point \((x_1, y_1)\) to a line \(Ax+By+C=0\) is \( d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \).
Step 3: Detailed Explanation
Finding the Center and Radius of the Circle:
The circle touches the x-axis at a distance of 3 units from the origin. Since the center is in the first quadrant, the point of contact must be \((3,0)\).
This means the x-coordinate of the center is 3. Let the center be \( (3,k) \). Since it touches the x-axis, the radius \(r\) is equal to the y-coordinate of the center, so \(r=k\). The center is \( (3,r) \).
Now, we use the information about the y-intercept. The y-axis is the line \(x=0\).
The perpendicular distance (\(d_y\)) from the center \( (3,r) \) to the y-axis is 3.
The length of the y-intercept is given by \( 2\sqrt{r^2 - d_y^2} \). We are given this length is \(6\sqrt{3}\).
\[ 2\sqrt{r^2 - 3^2} = 6\sqrt{3} \] \[ \sqrt{r^2 - 9} = 3\sqrt{3} \] Squaring both sides: \[ r^2 - 9 = (3\sqrt{3})^2 = 9 \times 3 = 27 \] \[ r^2 = 27 + 9 = 36 \] \[ r = 6 \quad (\text{since radius must be positive}) \] So, the circle's center is \( (3,6) \) and its radius is \( r=6 \).
The equation of the circle is \( (x-3)^2 + (y-6)^2 = 36 \).
Finding the Chord Length:
We need to find the length of the chord intercepted by this circle on the line \(x-y=3\), or \(x-y-3=0\).
First, calculate the perpendicular distance (\(d\)) from the center \( (3,6) \) to this line.
\[ d = \frac{|1(3) - 1(6) - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|3 - 6 - 3|}{\sqrt{1+1}} = \frac{|-6|}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \] Now, use the chord length formula: Length = \(2\sqrt{r^2 - d^2}\).
\[ \text{Chord Length} = 2\sqrt{6^2 - (3\sqrt{2})^2} \] \[ = 2\sqrt{36 - (9 \times 2)} = 2\sqrt{36 - 18} \] \[ = 2\sqrt{18} = 2(3\sqrt{2}) = 6\sqrt{2} \] Step 4: Final Answer
The length of the chord is \( 6\sqrt{2} \).