The line 2y+x=10 meets the circle x^2+y^2=65 at P and Q. Calculate the length of PQ?
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Hi Tae, if we were to draw both of these on a graph we would get a circle centered at the origin and a line that meets the circle at two points. So, to find the answer, we need to find the values of x and y for which the equations are equal. This turns into a simultaneous equation. When we find the final values, these will represent the two points (P,Q) that we then find the distance between. For clarity, I want to rewrite 2y + x = 10 as y = 5 - x/2. A: x^2 + y^2 = 65 B: y = 5 - x/2 Here is my general solution: x^2 + (5 - x/2)^2 = 65 (insert B into A) x^2 + 25 - 10x/2 + x^2/4 = 65 (expanded (5 - x/2)^2) (5x^2)/4 - 5x - 40 = 0 (simplify) 5x^2 - 20x - 160 = 0 (simplify more to get a quadratic equation, woo!) (5x - 40)(x + 4) = 0 (factorise) x = -4 and x = 8 (find possible solutions to equation) y = 7 and y = 1 (insert values of x from above into equation B) (-4, 7) and (8, 1) (now we have two co-ordinate sets) ((8 - (-4))^2 + (1 - 7)^2)^(1/2) (use the formula ((x2 - x1)^2 - (y2 - y1)^2)^(1/2) to find the distance between two points on a straight line) 180^(1/2) (simplify) There we have it Tae! I hope you find that helped :)
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