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Tangent drawn from the point P(4,0) to t...

Tangent drawn from the point `P(4,0)` to the circle `x^2+y^2=8` touches it at the point `A` in the first quadrant. Find the coordinates of another point `B` on the circle such that `A B=4` .

Text Solution

Verified by Experts

The correct Answer is:
`(2,-2)` or (-2,2)
`cos theta=(2sqrt(2))/(4)=(1)/(sqrt(2))`
`theta=45^(@)`
`:. A-= (2,2)`
Let `B -= (x_(1),y_(1))`

Given AB =4 . Therefore,
`(x_(1)-2)^(2)+(y_(1)-2)^(2)=16`
or `x_(1)^(2)+y_(1)^(2)-4x_(1)-4y_(1)=8`
Also, `x_(1)^(2)+y_(1)^(2)=8`
`:. x_(1)+y_(1)=0`
`:. 2x_(1)^(2)=8`
or `x_(1) +- 2`
`:. B -= (2,-2) ` or `(-2,2)`
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