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Question

# A and B are two points in the xy-plane, which are 2√2 unit distance apart and subtend an angle of 90o at the point C(1,2) on the line x−y+1=0, which is larger than any angle subtended by the line segment AB at any other point on the line. Find the equation(s) of the circle through the points A,B and C.

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Solution

## As shown in the figure, equation of line is L=x−y+1=0points A and B are endpoints of diameter as shown in figure. (Property of angle subtended in semicircle)Equation of family of circles touching a line L=0 at point (x1,y1) is,(x−x1)2+(y−y1)2+λL=0 where, λ = constant parameterHere, x1=1, y1=2∴(x−1)2+(y−2)2+λ(x−y+1)=0∴x2−2x+1+y2−4y+4+λx−λy+λ=0∴x2+y2+x(λ−2)+y(−λ−4)+(5+λ)=0∴x2+y2+2(λ−22)x+2(−λ−42)y+(5+λ)=0Comparing this equation with standard equation of circle i.e. x2+y2+2gx+2fy+c=0, we get,g=λ−22f=(−λ−42)c=5+λThus, radius of circle is given as,r=√g2+f2−c∴r=√(λ−22)2+(−(λ+42))2−(5+λ)∴r=√(λ−22)2+(λ+42)2−5−λ∴r=√λ2−4λ+44+λ2+8λ+164−5−λ∴r=√λ2−4λ+4+λ2+8λ+16−20−4λ4∴r=√2λ24∴r=√λ22∴r=±λ√2 (1)Now, AB is diameter and l(AB)=2√2 (given)∴d=2√2∴r=√2 (2)From (1) and (2),∴√2=±λ√2∴λ=±2Thus, equations of circles touching line L and passing through (1,2) are-1) (x−1)+(y−2)+2(x−y+1)=0∴x−1+y−2+2x−2y+2=0∴3x−y−1=02) (x−1)+(y−2)−2(x−y+1)=0x−1+y−2−2x+2y−2=0∴−x+3y−5=0∴x−3y+5=0

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