Question

The equation of a circle passing through the origin and making x-intercept 3 and y-intercept -5 is:

• A. \( x^2 + y^2 + 3x + 5y = 0 \)
• B. \( x^2 + y^2 + 3x - 5y = 0 \)
• C. \( x^2 + y^2 - 3x + 5y = 0 \)
• D. \( x^2 + y^2 - 3x - 5y = 0 \)

Hint:

For circles with known intercepts, use the intercepts to determine the values of \( g \) and \( f \) in the general equation of the circle.

Answer 1

The Correct Option is C

Step 1: General equation of the circle.
The general equation of a circle is: \[ x^2 + y^2 + 2gx + 2fy + c = 0 \] Since the circle passes through the origin, we substitute \( x = 0 \) and \( y = 0 \) into the equation: \[ 0 + 0 + 2g \cdot 0 + 2f \cdot 0 + c = 0 \quad \Rightarrow \quad c = 0 \] Thus, the equation becomes: \[ x^2 + y^2 + 2gx + 2fy = 0 \] Step 2: Apply the x- and y-intercepts.
We are given the x-intercept is 3 and the y-intercept is -5. For the x-intercept, set \( y = 0 \), and for the y-intercept, set \( x = 0 \). Substituting these values: For the x-intercept, \( y = 0 \): \[ 3^2 + 0^2 + 2g \cdot 3 + 2f \cdot 0 = 0 \quad \Rightarrow \quad 9 + 6g = 0 \quad \Rightarrow \quad g = -\frac{3}{2} \] For the y-intercept, \( x = 0 \): \[ 0^2 + (-5)^2 + 2g \cdot 0 + 2f \cdot (-5) = 0 \quad \Rightarrow \quad 25 - 10f = 0 \quad \Rightarrow \quad f = \frac{5}{2} \] Step 3: Final equation.
Substitute \( g = -\frac{3}{2} \) and \( f = \frac{5}{2} \) into the equation: \[ x^2 + y^2 - 3x + 5y = 0 \] Step 4: Conclusion.
Thus, the equation of the circle is \( \boxed{x^2 + y^2 - 3x + 5y = 0} \).