Question

Equation of the tangent to the circle, at the point $(1, - 1)$ , whose centre is the point of intersection of the straight lines $x - y = 1;2x + y = 3$ is
A) $x + 4y + 3 = 0$
B) $3x - y - 4 = 0$
C) $x - 3y - 4 = 0$
D) $4x + y - 3 = 0$

Answer 1

Hint: From the given two equations, find out the centre of the circle, then the paint is given on the circle you can find the radius of the circle by distance formula. We know the equation of circle is ${(x - \alpha )^2} + {(y - \beta )^2} = {r^2}$ where $(\alpha ,\beta )$ are the centre and r is the radius of the circle. Hence the equation of tangent is given by
$y = mx \pm a\sqrt {1 + {m^2}} $ where a represent radius this equation is valid for ${x^2} + {y^2} = {r^2}$.

Complete step-by-step answer:
Here , in this question , we are said to find the equation of tangent to the circle at point $(1, - 1)$ and it is given that centre is the intersection point of the straight lines $x - y = 1;2x + y = 3$
So,

Let O be the centre of circle having coordinates $(\alpha ,\beta )$
We are given ,$x - y = 1;2x + y = 3$
$
  x = 1 + y \\
  {\text{put in }}(2x + y = 3) \\
  2(1 + y) + y = 3 \\
  2 + 2y + y = 3 \\
  3y = 1 \\
  y = \dfrac{1}{3} \\
  so, \\
  x = 1 + \dfrac{1}{3} = \dfrac{4}{3} \\
 $
So, we got centre of the circle $(\dfrac{4}{3},\dfrac{1}{3})$
So, there is a point p from where we need to find the equation of the tangent.

Here when tangent is drawn from point p , then we know that tangent is always perpendicular to radius.
So, ${m_{\tan gent}}.{m_{radius}} = - 1$
(Slope of tangent )(slope of radius) $ = - 1$
$
  {m_{radius}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{\dfrac{1}{3} + 1}}{{\dfrac{4}{3} - 1}} = 4 \\
  {\text{hence, }}{m_{{\text{tangent}}}}.4 = - 1 \\
  {m_{{\text{tangent}}}} = - \dfrac{1}{4} \\
 $
And OP is the radius of circle
So,
OP will be
  $
   = \sqrt {{{({y_2} - {y_1})}^2} + {{({x_2} - {x_1})}^2}} \\
   = \sqrt {{{((\dfrac{1}{3}) + 1)}^2} + {{((\dfrac{4}{3}) - 1)}^2}} \\
   = \sqrt {{{(\dfrac{4}{3})}^2} + {{(\dfrac{1}{3})}^2}} \\
   = \sqrt {(\dfrac{{16}}{9}) + (\dfrac{1}{9})} \\
   = \sqrt {(\dfrac{{17}}{9})} \\
 $
We know the equation of tangent is given by
$y = mx \pm a\sqrt {1 + {m^2}} $
But this is valid for $({x^2} + {y^2} = {a^2})$ but here the circle is of form ${(x - \alpha )^2} + {(y - \beta )^2} = {r^2}$
So, equation of tangent is in the form of
$
  (y - \beta ) = m(x - \alpha ) \pm a\sqrt {(1 + {m^2})} \\
  {\text{here}};(\alpha ,\beta ) = {\text{centre}} \\
  a = {\text{radius}} \\
  m = {\text{slope(tangent)}} \\
  (y - \dfrac{1}{3}) = \dfrac{{ - 1}}{4}(x - \dfrac{4}{3}) \pm \sqrt {\dfrac{{17}}{9}} \sqrt {1 + {{(\dfrac{1}{4})}^2}} \\
  y - \dfrac{1}{3} = \dfrac{{ - x}}{4} + \dfrac{1}{3} \pm \sqrt {\dfrac{{17}}{9} \times (1 + \dfrac{1}{{16}})} \\
  y = \dfrac{{ - x}}{4} + \dfrac{2}{3} + \sqrt {\dfrac{{17 \times 17}}{{9 \times 16}}} \\
  y = - \dfrac{x}{4} + \dfrac{2}{3} \pm \dfrac{{17}}{{12}} \\
 $

Case I if we take +ve sign	Case II if we take –ve sign
$ y = \dfrac{{ - x}}{4} + \dfrac{2}{3} + \dfrac{{17}}{{12}} \\ y = \dfrac{{ - x}}{4} + \dfrac{{25}}{{12}} \\ y = \dfrac{{ - 3x + 25}}{{12}} \\ 12y + 3x = 25 \\ $	$ y = \dfrac{{ - x}}{4} + \dfrac{2}{3} - \dfrac{{17}}{{12}} \\ y = \dfrac{{ - x}}{4} - \dfrac{9}{{12}} \\ y = \dfrac{{ - x - 3}}{4} \\ 4y + x + 3 = 0 \\ $

Now we know that point P is on the tangent. So, it must satisfy both the given equations

$ 12y + 3x = 25 \\ 12( - 1) + 3(1) = 25 \\ - 12 + 3 = 25 \\ - 9 = 25 \\ $

$ 4y + x + 3 = 0 \\ 4( - 1) + 1 + 3 = 0 \\ - 4 + 4 = 0 \\ 0 = 0 \\ LHS = RHS \\ $

So, option A is correct.

Note: For a given circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ if we are given to find the equation of tangent at point P \[\left( {x,y} \right)\;\] lying on the circle then it is given by $x.{x_1} + y.{y_1} + g(x + {x_1}) + f(y + {y_1}) + c = 0$ here just replace
$
  {x^2} - > x.{x_1};{y^2} - > y.{y_1} \\
  x - > \dfrac{{x + {x_1}}}{2};y - > \dfrac{{y + {y_1}}}{2} \\
 $
We will get the desired answer.