Question
Find the equation to the circle which passes through the points $(1, 1) (2, 2)$ and whose radius is $1.$ Show that there are two such circles.
✓
Answer
Let $x^2 + y^2 + 2gx + 2fy + c = 0 ........(1)$
be the required cirde.
Now $(1)$ passes through $\text{P}=(1,\ 1)\ \&\ \theta(2,\ 2)$
$\therefore 1 + 1 + 2g + 2f + C = 0 ........... (2)$
$4 + 4 + 4g + 4f + c = 0 ............... (3)$
Also racius $= 1$
$\Rightarrow\sqrt{\text{g}^2+\text{f}^2-\text{c}}=1$
$\Rightarrow\text{g}^2+\text{f}^2-\text{c}1\ .......(4)$
from $(2)$ & $(4)$
$\text{g}+\text{f}+\frac{\text{c}}{2}=-1\ \&$
$\text{g}+\text{f}+\frac{\text{c}}{4}=-2$
on subtraction
and $\text{g}+\text{f}=-3\ ........(4)$
From (4) $\text{g}^2+\text{f}^2=5$
$\big\{\therefore(\text{g}+\text{f})^2=\text{g}^2+\text{f}^2+2\text{g}\text{f}\big\}$
$\therefore2\text{gf}=4$
$\Rightarrow9=5+2\text{gf}$
$\text{gf}=2$
so, $(\text{g}-\text{f})^2=\text{g}+\text{f})^2-4\text{gf}=9-8=1$
$\therefore\text{g}-\text{f}=\pm1\ .........\ (4)$
Solving $(5)$ & $(6)$ we get
$\text{g}=-1$ or $-2\ \&$
$\text{f}=-2$ or $-1$
Thus, required circle
$\text{x}^2+\text{y}^2-2\text{x}-4\text{y}+4=0$
$\text{x}^2+\text{y}^2-4\text{x}-2\text{y}+4=0$
be the required cirde.
Now $(1)$ passes through $\text{P}=(1,\ 1)\ \&\ \theta(2,\ 2)$
$\therefore 1 + 1 + 2g + 2f + C = 0 ........... (2)$
$4 + 4 + 4g + 4f + c = 0 ............... (3)$
Also racius $= 1$
$\Rightarrow\sqrt{\text{g}^2+\text{f}^2-\text{c}}=1$
$\Rightarrow\text{g}^2+\text{f}^2-\text{c}1\ .......(4)$
from $(2)$ & $(4)$
$\text{g}+\text{f}+\frac{\text{c}}{2}=-1\ \&$
$\text{g}+\text{f}+\frac{\text{c}}{4}=-2$
on subtraction
and $\text{g}+\text{f}=-3\ ........(4)$
From (4) $\text{g}^2+\text{f}^2=5$
$\big\{\therefore(\text{g}+\text{f})^2=\text{g}^2+\text{f}^2+2\text{g}\text{f}\big\}$
$\therefore2\text{gf}=4$
$\Rightarrow9=5+2\text{gf}$
$\text{gf}=2$
so, $(\text{g}-\text{f})^2=\text{g}+\text{f})^2-4\text{gf}=9-8=1$
$\therefore\text{g}-\text{f}=\pm1\ .........\ (4)$
Solving $(5)$ & $(6)$ we get
$\text{g}=-1$ or $-2\ \&$
$\text{f}=-2$ or $-1$
Thus, required circle
$\text{x}^2+\text{y}^2-2\text{x}-4\text{y}+4=0$
$\text{x}^2+\text{y}^2-4\text{x}-2\text{y}+4=0$
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