Write the standard form of the equation for the circle that passes through points (29, 1), (-19,1) and (-2, 18)

These are difficult.  You'll have 3 equations with 3 unknowns in the end to solve in order to get the coefficients right.  The standard form for a circle with this type of info is .  We have 3 points with different x and y coordinates that we will sub in to get these 3 equations we seek.  First point (29, 1): .  This simplifies down to 842+29D + E + F =0.  That's the first of 3 equations.  Next point (-19, 1):  which simplifies down to 362-19D+E+F=0.  That's the second equation.  Last point (-2, 18):  .  Which simplifies down to 328-2D+18E+F=0. That's the 3rd equation. In all of these move the constants over to the other side of the equals sign. 29D+E+F= -842; -19D+E+F= -362; -2D+18E+F= -328.  Take the first 2 equations and solve for D.  Do this multiplying one of them by -1 to get a set of equations that are 29D+E+F= -842 and 19D-E-F= 362.  Solve this by elimination and the E and the F cancel each other out leaving you with 48D= -480 and D = -10. Now take the second 2 equations and sub in the value for D you just found and work on eliminating either E or F.  -19(-10) +E+F= -362 which simplifies to 190+E+F= -362; -2(-10)+18E+F= -328 simplifies to 20+18E+F= -328.  Move the constants from the left to the right by subtraction to get a set of equations that is E+F= -552 and 18E+F= -348.Solve that first equation for E: E=-552-F and sub it into the second equatiion.  18(-552-F)+F= -348 and F= -564. Now let's go back to E+F= -552 and sub in our F of -564 to find that E = 12. Therefore, our equation, when all is said and done an hour later, is .  Not sure if that's the form your teacher wants it in, but that's standard. Ugh.

It is 5.8 you add all of the up to equat 180 then solve the equation