Question

Find the equation of a circle passing through point $(1,2)$ and $(3,4)$ and touching the line $3x+y-3=0$

Answer 1

Circle $(1,2)$ and $(3,4)-3x+y-3=0\Rightarrow (0,1)(1,0)$

Let $(x,y)$ be centre of circle $\Rightarrow$ Radius is equal both points.

$\sqrt{(x+1{)}^2+(y-2{)}^2}=\sqrt{(x-3{)}^2(y-2{)}^2}$....(1)

$x^2+1-2x+y^2+4-4y=x^2+9-6x+y^2+16-8y$

$-20+4x+4y=0\Rightarrow x+y=5\Rightarrow y=5-x$......(2)

distance from center of circle is same as radius.

$\frac{|3x+y-3|}{(3{)}^2+(1{)}^2}=\frac{|3x+5-x-3|}{\sqrt{10}}=\frac{|2x+2|}{\sqrt{10}}$....(3)

Substituting $\left(2\right)in\left(1\right)4$ equality to $3$

$\sqrt{(x-1{)}^2+(3-x{)}^2}=\frac{|2x+2|}{\sqrt{10}}\Rightarrow (x-1{)}^2+(3-x{)}^2=\frac{(2x+2{)}^2}{10}$

$x^2+1-2x+9+x^2-6x=4x^2+4\frac{x}{10}$

$16x^2-88x+96=0$

Solving form $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{88+\pm \sqrt{(88{)}^2-4\times 16\times 96}}{2\times 16}=\frac{88\pm 40}{32}$

$x_1=\frac{86+40}{32},x_2=\frac{88-40}{32}$

$x_1=4,x_2=1.5$

Substiting $x_{1}and{x}_2$ in (2) $y_1=1,y_2=3.5$

we get two circle with center $(4,1)$ and $(1.5,3.5)$

Radius of circle $C_1=(4.1{)}^2+(3-4{)}^2=\sqrt{10}$

$C_2=(1.5-1{)}^2+(3-1.5{)}^2=\sqrt{2.5}$

$C_1\Rightarrow (x-y{)}^2+(y-1{)}^2=10C_2\Rightarrow (x-1.5{)}^2+(y-3.5{)}^2=2.5$