Question

The point of contact of the line 2x+under root 5y-9=0 with the circle x~2+y`2=9 is

Answer 1

Answer:

9x+y-28=0

⇒y= -9x+28

put this value in circle equation

after solving

x=333

by putting this value in eq

y= 370

so answer is (333,370)

Step-by-step explanation:

Answer 2

The point of contact of the line $2x+\sqrt{5} y-9=0$ with the circle $x^{2} +y^{2} =9$ is $( 7,-\sqrt{5} )$.

Step-by-step explanation:

Given:

The line $2x+\sqrt{5} y-9=0$ make contact with the circle $x^{2} +y^{2} =9$.

To Find:

The point of contact of the line $2x+\sqrt{5} y-9=0$ with the circle $x^{2} +y^{2} =9$

Solution:

As given-The line $2x+\sqrt{5} y-9=0$ make contact with the circle $x^{2} +y^{2} =9$.

To find out  point of contact (x,y )  for this  it is required to solve both equations $2x+\sqrt{5} y-9=0$ and  $x^{2} +y^{2} =9$.

$2x+\sqrt{5} y-9=0$  ------ equation no.01.

$x^{2} +y^{2} =9$ ------------ equation no.02.±

Putting the value of x from equation no.01 in equation no.02. We get.

$(\frac{9-\sqrt{5}y }{2} )^{2} +y^{2} =9$

$\frac{(9-\sqrt{5} y )^{2} }{4} +y^{2} =9$

$(9-\sqrt{5} y )^{2} } +4y^{2} =36$

$9^{2} +(\sqrt{5} y)^{2}+2\times\ 9\times\sqrt{5} +4y^{2} =36$

$81+5y^{2} +18\sqrt{5} y +4y^{2} =36$

$9y^{2} +18\sqrt{5} y+45=0$

$y^{2} +2\sqrt{5} y+5=0$

$y=\frac{-2\sqrt{5}\± \sqrt{(2\sqrt{5} )^{2}-4\times 1\times5} }{2\times1}$

   $=\frac{-2\sqrt{5}\± \sqrt{20-20} }{2}$

  $=\frac{-2\sqrt{5}\± \sqrt{0} }{2}$

 $= -\sqrt{5}$

Putting the value of y in equation no.01,We get.

 $2x+\sqrt{5} (-\sqrt{5} )-9=0$

$2x-5-9=0$

$2x=14\\x=7$

Thus,the point of contact of the line $2x+\sqrt{5} y-9=0$ with the circle $x^{2} +y^{2} =9$ is $( 7,-\sqrt{5} )$.

PROJECT CODE#SPJ3