Question Detail

First divide by 2: x² + y² - x - (5/2)y + 3/2 = 0. Center: (1/2, 5/4). Check if (1,1) lies on circle: 2(1)²+2(1)²-2(1)-5(1)+3 = 2+2-2-5+3=0, yes. Equation of tangent at (1,1): For circle ax²+ay²+2gx+2fy+c=0, tangent is axx1+ayy1+g(x+x1)+f(y+y1)+c=0. Here a=2, 2g=-2→g=-1, 2f=-5→f=-2.5, c=3. Tangent: 2x(1)+2y(1)-1(x+1)-2.5(y+1)+3=0 → 2x+2y-x-1-2.5y-2.5+3=0 → x -0.5y -0.5=0 → multiply 2: 2x - y -1=0. Normal slope = -1/2 (perpendicular to tangent slope 2). Equation through (1,1): y-1 = (-1/2)(x-1) → 2y-2 = -x+1 → x+2y-3=0 → x+2y=3.