Find the equation of C2

Question

anonymous · Answer

attachment

anonymous · Answer

1.find equation of QS by$$ y-y1=\frac{ y2-y1 }{ x2-x1 }\left( x-x1 \right)$$
2.find perpendicular distance d from R to above line QS.
3. Find QS,QR,RS by distance formula $$\sqrt{\left( x2-x1 \right)^2+\left( y2-y1 \right)^2}$$
4.Area of triangle A$$=\frac{ 1 }{ 2 } *QS*d$$
5.Find perimeter p by p=QS+QR+RS
6. If r is the radius of inscribed circle then
$$A=\frac{ 1 }{ 2 }*p*r,r=\frac{ 2A }{ p }$$

anonymous · Answer

What is the equation of C2

anonymous · Answer

let O (x1,y1) be the centre of circle
then find perpendicular distance of O from lines QR,QS=r (already found)
solve these two equations for x1,y1
Equation of circle is $$\left( x-x1 \right)^2+\left( y-y1 \right)^2=r^2$$

anonymous · Answer

sorry I can't get the answer, can you show me step by step, thanks

mathmath333 · Answer

this one is correct



$\large\tt \color{black}{q=(-2,1),r=(2,5),s=(5,2)}$

$\large\tt \color{black}{qr=\sqrt{(-2-2)^2+(1-5)^2}}$

$\large\tt \color{black}{qr=4\sqrt{2}}$

$\large\tt \color{black}{rs=\sqrt{(2-5)^2+(5-2)^2}}$

$\large\tt \color{black}{rs=3\sqrt{2}}$

$\large\tt \color{black}{qs=\sqrt{(-2-5)^2+(1-2)^2}}$

$\large\tt \color{black}{qs=5\sqrt{2}}$

$\large\tt \color{black}{A=\dfrac{rs\times qr}{2}}$

$\large\tt \color{black}{A=\dfrac{ 3\sqrt2 \times 4\sqrt2}{2}}$

$\large\tt \color{black}{A=12}$

$\large\tt \color{black}{p=qr+rs+qs}$

$\large\tt \color{black}{p=12\sqrt{12}}$

from a.)

$\large\tt \color{black}{A=\dfrac{pr}{2}}$


$\large\tt \color{black}{r=\dfrac{2A}{p}}$

$\large\tt \color{black}{r=\sqrt2}$

let the centre of circle  be (h,k)

equation of line qr

$\large\tt \color{black}{y+2=\dfrac{5-1}{2+2}(x-1)}$

$\large\tt \color{black}{x-y-3=0}$

perpendicular distance from (h,k) to x-y-3=0 is r

$\large\tt \color{black}{r=\dfrac{|(h-k-3)|}{\sqrt{1^2+(-1)^2}}}$

$\large\tt \color{black}{\sqrt2=\dfrac{|(h-k-3)|}{\sqrt{1^2+(-1)^2}}}$

$\large\tt \color{black}{h-k-5=0-->a'}$

equation of line rs

$\large\tt \color{black}{y-5=\dfrac{5-2}{2-5}(x-2)}$

$\large\tt \color{black}{x+y-7}$

perpendicular distance from (h,k) to x+y-7=0 is r


$\large\tt \color{black}{r=\dfrac{|(h+k-7)|}{\sqrt{1^2+(1)^2}}}$

$\large\tt \color{black}{\sqrt2=\dfrac{|(h+k-7)|}{\sqrt{2}}}$

$\large\tt \color{black}{h+k-9=0}$


from a' and b'


$\large\tt \color{black}{(h,k)=(7,2)}$


equation of circle will be


$\large\tt \color{black}{(x-h)^2+(y-k)^2=r^2}$

$\large\tt \color{black}{(x-7)^2+(y-2)^2=2}$

anonymous · Answer

But the point O(7,2) is outside the circle and is not the centre of the circle

mathmath333 · Answer

yes , there must be some fault

mathmath333 · Answer

http://www.wolframalpha.com/input/?i=plot+%28x-2%29%5E2%2B%28y-3%29%5E2%3D2+%2Cx-y%3D-3%2Cx%2By%3D7%2Cx-7y%3D-9

mathmath333 · Answer

$p=12\sqrt{2}$

mathmath333 · Answer

equation of line qr

$$x-y=-3$$

mathmath333 · Answer

$$\sqrt{2}=\dfrac{ h-k+3 }{ \sqrt2 }$$

mathmath333 · Answer

$$h-k+1=0$$

mathmath333 · Answer

$$h+k=5$$ it should be this

mathmath333 · Answer

to get (2,3)

anonymous · Answer

is that any other method to solve this question, not using the perpendicular distance formula