Question

Find the equation of the circle passing the point (1, 2) and touching the line 2x + y - 1 = 0 at the point (1, -1).

• A. $x^2+y^2+8x+y+y=0$
• B. $x^2+y^2-4x-2y+5=0$
• C. $x^2+y^2-8x-y+5=0$
• D. $x^2+y^{2}4x+2y-5=0$

Answer 1

The correct option is C

$x^2+y^2-8x-y+5=0$

Here, equation of family of circle is

$s+\lambda L=0$

$(x-1{)}^2+(y+1{)}^2+\lambda (2x+y-1)=0$ .........(1)

Since, It passes through the point (1,2)

Then

$(1-1{)}^2+(2+1{)}^2+\lambda (2\times 1+2-1)=0$

$9+\lambda \left(3\right)=0$

$\lambda =\frac{-9}{3}=-3$

Substituting the value of $\lambda$ in equation (1)

$(x-1{)}^2+(y+1{)}^2+(-3)(2x+y-1)=0$

$x^2+y^2-8x-y+5=0$

Equation of circle is

$x^2+y^2-8x-y+5=0$

Option c is correct