Question Detail

First divide by 2: x²+y²-x-(5/2)y+3/2=0. Center: (-g,-f) with g=-1/2, f=-5/4, so center (1/2,5/4). Radius vector from center to point (1,1): (1-1/2, 1-5/4) = (1/2, -1/4). Slope = (-1/4)/(1/2) = -1/2. Normal slope is perpendicular = 2. Equation through (1,1): y-1=2(x-1) → y-1=2x-2 → 2x-y-1=0. Not matching. For normal at (1,1), check if (1,1) lies on circle: 2+2-2-5+3=0, yes. Different approach: For circle ax²+ay²+2gx+2fy+c=0, equation of tangent at (x1,y1) 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, same as before. Normal is perpendicular: x+2y+c=0. Pass through (1,1): 1+2+c=0 → c=-3, so x+2y-3=0 → x+2y=3.