The centre of the circle passing through origin and making intercepts 8 and -4 on x and y-axes respectively is

Options

A. (4,-2)
B. (-2,4)
C. (8,-4)
D. both 1 and 2

Answer

To find the center of the circle that passes through the origin and makes intercepts of 8 on the x-axis and -4 on the y-axis, we can follow these steps:

### Step 1: Understanding the Circle's Equation
The general equation of a circle with center (a, b) is given by:
\[ (x - a)^2 + (y - b)^2 = r^2 \]
where (a, b) is the center of the circle and r is the radius.

### Step 2: Circle Passing Through the Origin
Since the circle passes through the origin (0, 0), we can substitute these coordinates into the circle's equation:
\[ (0 - a)^2 + (0 - b)^2 = r^2 \]
This simplifies to:
\[ a^2 + b^2 = r^2 \quad \text{(Equation 1)} \]

### Step 3: Circle's Intercepts
The circle makes intercepts of 8 on the x-axis and -4 on the y-axis. The x-intercept means that when y = 0, x = 8. Therefore, the point (8, 0) lies on the circle.

### Step 4: Substitute the x-intercept into the Circle's Equation
Substituting the point (8, 0) into the circle's equation:
\[ (8 - a)^2 + (0 - b)^2 = r^2 \]
Expanding this, we get:
\[ (8 - a)^2 + b^2 = r^2 \]
This expands to:
\[ 64 - 16a + a^2 + b^2 = r^2 \]

### Step 5: Equating the Two Expressions for r²
From Equation 1, we know that \( r^2 = a^2 + b^2 \). Therefore, we can set the two expressions for \( r^2 \) equal to each other:
\[ 64 - 16a + a^2 + b^2 = a^2 + b^2 \]
Cancelling \( a^2 + b^2 \) from both sides gives:
\[ 64 - 16a = 0 \]
Solving for a:
\[ 16a = 64 \implies a = 4 \]

### Step 6: Using the y-intercept
The circle also makes an intercept of -4 on the y-axis, meaning the point (0, -4) lies on the circle. Substituting this point into the circle's equation:
\[ (0 - a)^2 + (-4 - b)^2 = r^2 \]
Substituting \( a = 4 \):
\[ (0 - 4)^2 + (-4 - b)^2 = r^2 \]
This simplifies to:
\[ 16 + (-4 - b)^2 = r^2 \]
Using \( r^2 = a^2 + b^2 \) again:
\[ 16 + (b + 4)^2 = 16 + b^2 \]
Cancelling \( 16 \) from both sides gives:
\[ (b + 4)^2 = b^2 \]
Expanding the left side:
\[ b^2 + 8b + 16 = b^2 \]
Cancelling \( b^2 \) gives:
\[ 8b + 16 = 0 \]
Solving for b:
\[ 8b = -16 \implies b = -2 \]

### Step 7: Conclusion
Thus, the center of the circle is:
\[ (a, b) = (4, -2) \]

### Final Answer
The center of the circle is \( (4, -2) \).

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