Question

Find the equation of a circle $S(x,y)=0,S(2,3)=16$ which touches the line $3x+4y-7=0$ at $(1,1)$

Answer 1

Any circle which touches $3x+4y–7=0$ at $(1,1)$ will be of the form

$S(x,y)={(x–1)}^2+{(y–1)}^2+\lambda (3x+4y–7)=0$

Since $S(2,3)=16$

$\Rightarrow {(2–1)}^2+{(3–1)}^2+\lambda (3\times 2+4\times 3–7)=0$

$\Rightarrow 1+4+\lambda (6+12–7)=16$

$\Rightarrow 11\lambda =16-5=11$

$\therefore \lambda =1$

so the required circle is ${(x–1)}^2+{(y–1)}^2+(3x+4y–7)=0$

or $x^2-2x+1+y^2-2y+1+3x+4y–7=0$

or $x^2+y^2+x+2y-5=0$