Question

If the segments of the straight lines \[ x+y=6 \] and \[ x+2y=4 \] are two diameters of a circle passing through \((6,2)\), then the equation of that circle is:

• A. \(x^2+y^2-2x-4y-20=0\)
• B. \(x^2+y^2+6x-4y-68=0\)
• C. \(x^2+y^2-16x+4y+48=0\)
• D. \(x^2+y^2+2x-10y-32=0\)

Hint:

If two lines are diameters of a circle, their point of intersection is the centre of the circle.

Answer 1

The Correct Option is C

Step 1: Locate the centre as the intersection of the two diameters.
The diameters are given by x + y = 6 and x + 2y = 4. Subtracting the first from the second yields y = -2. Substituting back gives x - 2 = 6 → x = 8. The centre is (8, -2).

Step 2: Determine the radius using the point on the circle.
The circle passes through (6, 2). The squared radius is r² = (6 - 8)² + (2 - (-2))² = (-2)² + 4² = 4 + 16 = 20.

Step 3: Write the equation in standard and then expanded form.
With centre (8, -2) and r² = 20: (x - 8)² + (y + 2)² = 20. Expanding: x² - 16x + 64 + y² + 4y + 4 = 20 → x² + y² - 16x + 4y + 48 = 0.

Step 4: Final conclusion.
The equation of the circle is x² + y² - 16x + 4y + 48 = 0.