The general solution of the differential equa | vedclass

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(B) The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$. Given the center is $A(-1, 2)$,the equatio

Vedclass · Accepted Answer

$x^2 + y^2 + 2x - 4y + c = 0$

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(B) The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$. Given the center is $A(-1, 2)$,the equation becomes $(x - (-1))^2 + (y - 2)^2 = r^2$. Expanding this,we get $(x + 1)^2 + (y - 2)^2 = r^2$. $x^2 + 2x + 1 + y^2 - 4y + 4 = r^2$. $x^2 + y^2 + 2x - 4y + 5 - r^2 = 0$. Let $c = 5 - r^2$,where $c$ is an arbitrary constant. Thus,the general solution is $x^2 + y^2 + 2x - 4y + c = 0$.

The general solution of the differential equation of all circles having center at $A(-1, 2)$ is $ . . . . . . $.

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