please find at least in an hour.....
in Calculus Answers by Level 1 User (200 points)

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Only just seen your question, sorry. Complete the squares for x and y: 9(x^2+6x+9)+9(y^2-4y+4)-108-(81+36)=0⇒9(x+3)^2+9(y-2)^2-108-117=0⇒9(x+3)^2+9(y-2)^2=225. [To decide how to complete the squares, we take 9 outside the brackets and divide the x^2 and y^2 terms by 9 so that leaves x^2 and y^2 inside the brackets. Then we halve the coefficients of the x and y terms and square them to find out the constants squared. Half of 6 is 3 which squared is 9; half of 4 is 2 which squared is 4. Having completed the squares we have to subtract the constants we just added so as to maintain a balanced equation. We added 9 and 4 inside the brackets, now we must multiply the sum 13 by 9 from outside the brackets. So we subtract 117.] We can divide the equation throughout by 9: (x+3)^2+(y-2)^2=25. From this the h and k values are -3 and 2, putting the centre of the circle at (-3,2) and the radius is sqrt(25)=5. To draw the circle mark the point (-3,2) and set your compasses to 5 units. The graph needs to extend to -8 to +2 on the x axis and -3 to +7 on the y axis so that the circle fits on to the graph.

by Top Rated User (1.0m points)

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