The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from the positive parts of the x-axis and y-axis respectively is `(x-(5)/(2))^(2) =(y-3)^(2)=lambda`, where `lambda` is

Options

A. `(61)/(4)`
B. `(4)/(6)`
C. `(1)/(4)`
D. 0

Answer

To find the value of \(\lambda\) in the equation of the circle that passes through the origin and cuts off intercepts of 5 and 6 from the positive parts of the x-axis and y-axis respectively, we can follow these steps:

### Step 1: Identify the center of the circle
The intercepts on the x-axis and y-axis are given as 5 and 6 respectively. The center of the circle can be determined from these intercepts. The center \(C\) of the circle will be at the point \((\frac{5}{2}, 3)\).

### Step 2: Calculate the radius of the circle
The radius \(r\) of the circle can be calculated using the distance from the center to the origin (0, 0). The distance formula gives us:
\[ r = \sqrt{\left(\frac{5}{2} - 0\right)^2 + (3 - 0)^2} \]
Calculating this:
\[ r = \sqrt{\left(\frac{5}{2}\right)^2 + 3^2} = \sqrt{\frac{25}{4} + 9} = \sqrt{\frac{25}{4} + \frac{36}{4}} = \sqrt{\frac{61}{4}} = \frac{\sqrt{61}}{2} \]

### Step 3: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = \frac{5}{2}\), \(k = 3\), and \(r = \frac{\sqrt{61}}{2}\):
\[ \left(x - \frac{5}{2}\right)^2 + (y - 3)^2 = \left(\frac{\sqrt{61}}{2}\right)^2 \]
Calculating \(r^2\):
\[ r^2 = \left(\frac{\sqrt{61}}{2}\right)^2 = \frac{61}{4} \]
Thus, the equation becomes:
\[ \left(x - \frac{5}{2}\right)^2 + (y - 3)^2 = \frac{61}{4} \]

### Step 4: Identify \(\lambda\)
From the equation of the circle, we can see that \(\lambda\) corresponds to the right-hand side of the equation:
\[ \lambda = \frac{61}{4} \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \lambda = \frac{61}{4} \]

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