Q.Find the equation of the circle which passes through the points $(2,3)$ and $(4,5)$ and the centre lies on the straight line $y-4x+3=0$. (A) $x^2+y^2-4x-10y+25=0$ (B) $x^2+y^2-4x-10y-25=0$ (C) $x^2+y^2-4x+10y-25=0$ (D) $x^2+y^2+4x-10y+25=0$

Match the centre $\left(-\tfrac a2,-\tfrac b2\right)=(1,-2)$ to get $a=-2,\ b=4$, then radius $=\sqrt{1+4+4}=3$.

Compare with the general circle $x^2+y^2+2gx+2fy+c=0$ whose centre is $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$. Here $2g=a$ and $2f=b$, so the centre is $\left(-\tfrac a2,-\tfrac b2\right)$.

Given centre $(1,-2)$:

$-\frac a2 = 1 \Rightarrow a=-2,\qquad -\frac b2 = -2 \Rightarrow b=4$

With $g=-1,\ f=2,\ c=-4$:

$r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+2^2-(-4)} = \sqrt{1+4+4} = \sqrt{9} = 3$

Tangency to the x-axis and y-axis both require the center's distance to each axis to equal the radius (1); only the circle centered at $(-1,-1)$ satisfies both.

A circle with center $(h,k)$ and radius $r$ is tangent to the x-axis when $|k|=r$, and tangent to the y-axis when $|h|=r$. Here $r=1$, so we need $|h|=1$ and $|k|=1$, i.e. center at $(\pm1,\pm1)$.

1. $(x+1)^2+(y+1)^2=1$: center $(-1,-1)$, distances to both axes $=1=r$. Tangent to both. Valid.
2. $x^2+(y-1)^2=1$: center $(0,1)$, distance to y-axis $=0\ne1$ — the circle crosses the y-axis, not tangent. Invalid.
3. $(x-1)^2+y^2=1$: center $(1,0)$, distance to x-axis $=0\ne1$ — crosses the x-axis. Invalid.
4. $x^2+y^2=1$: center $(0,0)$, distance to both axes $=0\ne1$ — passes through both axes. Invalid.