Question

The general solution of the differential equation of all circles having center at A (-1,2) is?
A.$x^2+y^2+x-2y+c=0$
B.$x^2+y^2-2x+4y+c=0$
C.$x^2+y^2-x+2y+c=0$
D.$x^2+y^2+2x-4y+c=0$

Answer 1

The general form: $x^2+y^2+Dx+Ey+F =0$, where D, E, F are constants. If the equation of a circle is in the standard form, we can easily identify the center of the circle, (h, k), and the radius, r.

Complete step-by-step answer:
Let D(x) is the differential equation of all circles having Centre at A (−1, 2).
Then, by the very definition of D(x),
Any solution to D(x) will be a circle having Centre at A (−1, 2).
Centre of the circle is the point where the radius starts. Its coordinates are always (0, 0)
So, any solution to D will be of the form:
$(x+1)^2+(y-2)^2=k$
Now we expanding the terms and keep everything on the LHS,
That is, any solution to D will be of the form:
$x^2+y^2+2x-4y+c=0$
Where, c=5−k.
The in regards to the equation of circle having Centre at A (-1, 2) the answer is option (d).

Note: Students need to keep in mind that the Centre of circle with respect to the circle is at A (-1, 2). From there we put the values of (x, y) into the equation of the circle and find out the answer and the value of the constant term c