Original post by zezno
thanks again!
A circle of centre (a,b) and radius r has the equation
(x-a)^2 + (y-b)^2 = r^2
You can eliminate several wrong answers on the basis that r^2 must be positive, it has to x^2 plus y^2 and you won't have an xy term.
You need to complete the square for x and y terms in c and d to get them into the above format to find the right answer.
Edit:beaten to it
Posted from TSR Mobile
(edited 7 years ago)
Original post by gdunne42
A circle of centre (a,b) and radius r has the equation
(x-a)^2 + (y-b)^2 = r^2
You can eliminate several wrong answers on the basis that r^2 must be positive, it has to x^2 plus y^2 and you won't have an xy term.
You need to complete the square on c and d to get them into the above format to find the right answer.
Edit:beaten to it
Posted from TSR Mobile
(x-a)^2 + (y-b)^2 = r^2
You can eliminate several wrong answers on the basis that r^2 must be positive, it has to x^2 plus y^2 and you won't have an xy term.
You need to complete the square on c and d to get them into the above format to find the right answer.
Edit:beaten to it
Posted from TSR Mobile
you're beautiful
Original post by zezno
you're beautiful
You're welcome....
B_9710 was faster
Posted from TSR Mobile
Original post by gdunne42
@B_9710
any idea for this one? I managed to do Q2 and Q3 but a bit confused on this one. Thanks
Original post by zezno
For the first part sub x=2 into the equation and then solve for y.
Then you have the coordinates of these points. You can find the gradient of the line between the centre of the circle bs the 2 points. Then remember that the tangent to a circle is perpendicular to the radius.
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