Question

Point of contact of line y is equal to x + 4 root 2 to the circle x square + y square equal to 16

Answer 1

Answer:

$\text{The point of contact is }\bf\,(-2\sqrt{2},2\sqrt{2})$

Step-by-step explanation:

Given line is $y=x+4\sqrt{2}$....(1)

Given circle is $x^2+y^2=16$....(2)

The point conctact is found by solving equations (1) and (2)

using (2) in (1)

$x^2+(x+4\sqrt{2})^2=16$

$x^2+x^2+8\sqrt{2}x+32-16=0$

$2x^2+8\sqrt{2}x+16=0$

$x^2+4\sqrt{2}x+8=0$

$(x+2\sqrt{2})^2=0$

$x+2\sqrt{2}=0$

$\implies\boxed{\bf\,x=-2\sqrt{2}}$

put $x=-2\sqrt{2}$ in (1) we get

$y=-2\sqrt{2}+4\sqrt{2}$

$\implies\boxed{\bf\,y=2\sqrt{2}}$

$\therefore\text{The point of contact is }\bf\,(-2\sqrt{2},2\sqrt{2})$

Answer 2

Step-by-step explanation:

The point of contact is (−22,22)

Step-by-step explanation:

Given line is y=x+4\sqrt{2}y=x+42 ....(1)

Given circle is x^2+y^2=16x2+y2=16 ....(2)

The point conctact is found by solving equations (1) and (2)

using (2) in (1)

x^2+(x+4\sqrt{2})^2=16x2+(x+42)2=16

x^2+x^2+8\sqrt{2}x+32-16=0x2+x2+82x+32−16=0

2x^2+8\sqrt{2}x+16=02x2+82x+16=0

x^2+4\sqrt{2}x+8=0x2+42x+8=0

(x+2\sqrt{2})^2=0(x+22)2=0

x+2\sqrt{2}=0x+22=0

\implies\boxed{\bf\,x=-2\sqrt{2}}⟹x=−22

put x=-2\sqrt{2}x=−22 in (1) we get

y=-2\sqrt{2}+4\sqrt{2}y=−22+42

\implies\boxed{\bf\,y=2\sqrt{2}}⟹y=22

\therefore\text{The point of contact is }\bf\,(-2\sqrt{2},2\sqrt{2})∴The point of contact is (−22,22)