Fixed distance of any point on the circle from O

$=R \equiv Radius$

Fixed point O

$\equiv$ Centre of the circle.

A chord passing through the centre is called a Diameter.

slope $=\frac{2-1}{1-0}-1$

$\frac{y-1}{x-0}=1$

$y=x+1$

$8 \neq 7$

CENTRE : (h, K), \quad \quad RADIUS : R

$\triangle OPQ$

Pythagoras theorem

$(OP)^2 = (OQ)^2 + (PQ)^2$

$R^2 = (x-h)^2 + (y-K)^2$

CENTRE : (1, 2)

RADIUS : 5 = R

$(x,y)=(-1,3)$

$25 \neq 4 + 1$

CENTRE - RADIUS FORM

CENTRE: $(3,7)$

RADIUS: 8

Circle with centre (h, k) and Radius R

$OP=R$

$OQ=R \cos \theta$

$QP=R \sin \theta$

$x=h+R \cos \theta$

$y = k + R \sin \theta$

$(x-h)^2=R^2 \cos^2 \theta$

$(y-k)^2=R^2 \sin^2 \theta$

Circle: $\{(x, y) \mid x = h + R \cos \theta,\\ \quad \quad \quad \quad \quad\quad y = k + R \sin \theta\}$

Circle = (x, y) | $x=h+R \cos \theta$,

$\quad \quad \quad \quad y = k + R \sin \theta$

$\theta \in (0, 2 \pi)$

1) If a point (x, y) belongs to a circle having centre (h, K) and radius R,

than $x = h + R \cos \theta, y = k + R \sin \theta$

for some $\theta \in (0,2 \pi)$

2) For any angle $\theta \in (0, 2 \pi)$ the points $(h + R \cos \theta, k + R \sin \theta)$ belongs to the circle having centre (h, K) and radius R.

$(x-h)^2 + (y-k)^2 = R^2$

$x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = R^2$

$[x^2 + y^2 - 2hx - 2ky + ck^2 + h^2 - R^2] = 0$

$\rightarrow x^2 + y^2 + 2gx + 2fy + c = 0$

$ax^2 + by^2 + cxy + 2dx + 2ey + f = 0$

$\rightarrow$ general $2^{nd}$ degree equation

$a=b$ and $c=0$

$x^2 + y^2 + \frac{2d}{a} x + \frac{2 e}{a} y+\frac{f}{a} = 0$

$\rightarrow x^2 + y^2 + 2gx + 2fy + c = 0$

$(x^2 + 2gx + g^2) + (y^2 + 2fy + f^2) + c - g^2 - f^2 = 0$

$(x + g)^2 + (y + f)^2 = (g^2 + f^2 - c)$

$(x - h)^2 + (y - k)^2 = R^2 \geq 0$

$g^2 + f^2 - c \geq 0$

$R = \sqrt{g^2 + f^2 - c}$

$g^2 + f^2 - c \geq 0$, in which case the radius of the circle is

$R=\sqrt{g^2 + f^2 - c}$ and the centre of the circle is

$(h, k) = (-g, -f)$

Thank you