Circles Test 11

Fact: equation of circle with centre $$(h,k)$$ and radius $$r$$ is given by 

The equation of circle is $$x^{ 2 }+y^{ 2 }-6x+4y-12=0$$

• The radius of given circle is $$\sqrt{11} = 3.32$$

$$3x+4y+l=0\\ |OP|=\cfrac { -3g-4f+l }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } }  } \\ =\cfrac { -3g-4f+l }{ 5 } \\ |OD|=\cfrac { -6g-8f+m }{ 10 } \\ |OP|=|OD|\\ \cfrac { -3g-4f+l }{ 5 } =\cfrac { -6g-8f+m }{ 10 } \\ -6g-8f+2l=-6g-8f+m\\ 2l-m\\ 2l=m\\ 3a+4y=-l\times 2\\ 6x+8y=-2l\times 1\\ 6x+8y=-2l\\ 6x+8y=-2l$$

Given line is
$$(x-2)cos \,\theta +(y-2)sin\, \theta =1$$
$$\Rightarrow (x-2)cos \theta +(y-2)sin\theta=cos^2 \, \theta+sin^2\, \theta$$
On compaining we get,
$$(x-2) = cos\theta $$ ..... $$(i)$$
$$(y-2) = sin \theta$$ ..... $$(ii)$$
On squaring and then adding Eqs. $$(i)$$ and $$(ii),$$ we get
$$(x-2)^2+(y-2)^2=cos^2\theta +sin^2 \theta$$
$$\Rightarrow (x-2)^2+(y-2)^2=1$$
$$\Rightarrow x^2+y^2-4x-4y+7=0$$

If the smallest circle is drawn through the points $$(2,2)$$ and $$(3,3)$$ then, the point have to be the opposite end of the diameter of the circle.
Therefore, the centre of the circle is $$C=(\dfrac{5}{2},\dfrac{5}{2})$$.
The radius of the circle is $$R=\dfrac{\sqrt{2}}{2}$$ ....(using distance formula)
Hence equation of the circle is
$$(x-\dfrac{5}{2})^{2}+(y-\dfrac{5}{2})^{2}=\dfrac{1}{2}$$
$$x^{2}+y^{2}-5x-5y+\dfrac{50}{4}=\dfrac{1}{2}$$
Or
$$x^{2}+y^{2}-5x-5y+\dfrac{25}{2}-\dfrac{1}{2}=0$$
Or
$$x^{2}+y^{2}-5x-5y+12=0$$.

From the figure, it is clear that the centre of the circle will lie on either $$x$$ or $$y$$ axis.

Let us take the case when the centre lies on $$x$$ axis and let it be $$B(h,0)$$

Given $$OA=a$$

$$\Rightarrow OC=BC=\dfrac{a}{2}$$

Now in $$\triangle OBC$$ $$,$$ $${ OB }^{ 2 }={ OC }^{ 2 }+{ BC }^{ 2 }$$

$$\Rightarrow { OB }^{ 2 }={ \left( \dfrac { a }{ 2 }  \right)  }^{ 2 }+{ \left( \dfrac { a }{ 2 }  \right)  }^{ 2 }$$

$$ \Rightarrow { h }^{ 2 }=\dfrac { { a }^{ 2 } }{ 2 } $$

$$ \Rightarrow h=\pm \dfrac { a }{ \sqrt { 2 }  } $$

So the centre of the circle is $$\left( \pm \dfrac { a }{ \sqrt { 2 }  } ,0 \right) $$ and radius is $$\dfrac { a }{ \sqrt { 2 }  } $$ and its equation is 

$${ \left( x\pm \dfrac { a }{ \sqrt { 2 }  }  \right)  }^{ 2 }+{ (y-0) }^{ 2 }={ \left( \dfrac { a }{ \sqrt { 2 }  }  \right)  }^{ 2 }$$

$$ { x }^{ 2 }+\dfrac { { a }^{ 2 } }{ 2 } \pm \sqrt { 2 } ax+{ y }^{ 2 }=\dfrac { { a }^{ 2 } }{ 2 } $$

$$ { x }^{ 2 }\pm \sqrt { 2 } ax+{ y }^{ 2 }=0$$

Similarly when centre lies on $$y$$ axis centre will be $$\left( 0,\pm \dfrac { a }{ \sqrt { 2 }  }  \right) $$ and radius $$\dfrac { a }{ \sqrt { 2 }  } $$ and its equation will be

$${ x }^{ 2 }\pm \sqrt { 2 } ay+{ y }^{ 2 }=0$$

Equation of circle which touches the $$y$$-axis at origin is 

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