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Grade: 12th pass

                        

Please answer me I have attached photo with it .It make lot of cinfusion

2 years ago

Answers : (1)

Arun
24737 Points
							
Given circle x²-y²-2x+4y-3 = 0Convert into complete squarex²-2x-y²+4y=3(x²-2x) - (y²-4y) = 3(x²-2x+1) - (y²-4y+4) = 3+1-4(x-1)² - (y-2)² = 0a2 – b2 = (a + b)(a - b)[(x-1)-(y-2)] [(x-1)+(y-2)] = 0[x-1-y+2] [x-1+y-2] = 0(x-y+1)(x+y-3) = 0x-y+1 = 0 and x+y-3 = 0 -y = -x-1 y = x+1 and y = -x+3Thease are the equations of the two lines.Since two diameters of the same circle must intersect at the center of the circle. The center of the circle must be the intersection of those two lines.Solve the equations of the two lines to find their point of intersection.y = -x+3y = x+1eqating we get -x+3 = x+1 -2x = -2 x = 1Substitutey = 1+1y=2So that intersect at (1,2) is the center and the radius is the distance from the center to the point on the circle (1,1) is 1 unit. So the equation of the circle is(x-h)² + (y-k)² = r² (x-1)² + (y-2)² = 1x²-2x+1+y²-2y+4 = 1 or x²+y²-2x-2y+3 = 0
2 years ago
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