Circles Test 3

According to question the centre of the circle is the intersection of lines $$L_1 :2x+y=3$$ and $$L_2:3x+2y=5$$

Consider the given circle equation.

$${{x}^{2}}+{{y}^{2}}-4x-6y+4=0$$            …… (1)

 

We know that the general equation of circle,

$${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$$       …… (2)

 

On comparing equation (1) and (2), we get

$$ 2g=-4\Rightarrow g=-2 $$

$$ 2f=-6\Rightarrow f=-3 $$

$$ c=4 $$

 

So, the centre of the circle

$$ C=\left( -g,-f \right) $$

$$ C=\left( 2,3 \right) $$

 

We know that the radius of circle,

$$r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$$

$$ r=\sqrt{{{2}^{2}}+{{3}^{2}}-4} $$

$$ r=\sqrt{4+9-4} $$

$$ r=\sqrt{9}=3 $$

 

So, the diameter of this circle

$$d=2r=6$$

 

Hence, this is the answer.

Let $$\left( {x,y} \right)$$ be point then ATQ

                $$\sqrt {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2} = 5} $$

                $${\left( {x - 2} \right)^2} + {\left( {y - 3} \right)^2} = 25$$

                $${x^2} + {y^2} + 4 + 9 - 4x - 6y = 25$$

                $$ \Rightarrow {x^2} + {y^2} - 4x - 6y - 12 = 0$$

The center radius form equation of circle is $$(x-h)^2+(y-k)^2=r^2$$, where $$(h,k)$$ is the center and $$r$$ is the radius.

To find circumcentre we write the equation of perpendicular bisectors of two sides and find their intersection,

3x-y-3=0 and 6x+8y-21=0

Their intersection point is ($$\dfrac{3}{2}$$,$$\dfrac{3}{2}$$)

Radius of circumcircle = Distance of ($$\dfrac{3}{2}$$,$$\dfrac{3}{2}$$)

from (2,-2) or any other vertex = $$\dfrac{5}{\sqrt{2}}$$

So equation of circle = ($$x-\frac{3}{2})^2$$ +($$y-\frac{3}{2})^2$$=$$\dfrac{25}{2}$$ which corresponds to B option .

$$9{ x }^{ 2 }-k{ x }^{ 2 }+{ y }^{ 2 }+k{ y }^{ 2 }+2kx=0\\ { x }^{ 2 }(9-k)+{ y }^{ 2 }(1+k)+2kx=0$$

$$\begin{array}{l} The\, equation\, of\, line\, pas\sin  g\, through\, centre \\ \vec { i } +2\vec { k } \, and\, normal\, to\, the\, given\, plane \\ \vec { r } =\vec { j } +2\vec { k } +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k }  } \right) \, \, \, \, \, \, -----\left( 1 \right)  \\ This\, plane\, meets\, the\, plane\, at\, a\, po{ { int } }\, for\, which \\ \left[ { \left( { \vec { j } +2\vec { k }  } \right) +\lambda \left( { \vec { i } +2\vec { j } +2\vec { k }  } \right)  } \right] \left( { \vec { i } +2\vec { j } +2\vec { k }  } \right) =15 \\ \Rightarrow 6+9\lambda =15 \\ \Rightarrow 9\lambda =9 \\ \therefore \lambda =1 \\ substituting\, value\, of\, \lambda \, in\, eq{ u^{ n } }\left( 1 \right)  \\ \vec { r } =\vec { j } +2\vec { k } +\vec { i } +2\vec { j } +2\vec { k }  \\ \Rightarrow \vec { r } =\vec { i } +3\vec { j } +4\vec { k }  \\ \therefore Centre\, is\, \left( { 1,\, 3,\, 4 } \right)  \end{array}$$