What is the equation for a circle with a center at (5,-2) and passes through (4,0)?
1 answer:
Step-by-step explanation:
since you said equation I'm assuming it's in terms of x and y Just as a general equation
since the centre of the circle isn't at 0, you have to take that into account using the formula above. solve for the radius of the circle by substituting (4,0) into the formula and get your r value to replace in the original formula
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Answer:
x = 3 + √6 ; x = 3 - √6 ; ;
Step-by-step explanation:
Relation given in the question:
(x² − 6x +3)(2x² − 4x − 7) = 0
Now,
for the above relation to be true the following condition must be followed:
Either (x² − 6x +3) = 0 ............(1)
or
(2x² − 4x − 7) = 0 ..........(2)
now considering the equation (1)
(x² − 6x +3) = 0
the roots can be found out as:
for the equation ax² + bx + c = 0
thus,
the roots are
or
or
and, x =
or
and, x =
or
x = 3 + √6 and x = 3 - √6
similarly for (2x² − 4x − 7) = 0.
we have
the roots are
or
or
and, x =
or
and, x =
or
and, x =
or
and,
Hence, the possible roots are
x = 3 + √6 ; x = 3 - √6 ; ;