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The equation of the triangle is x^2 + y^2 = 16. The circle does not pass through the given point (2, 2) as 2^2 + 2^2 = 4 + 4 = 8 not 16.
To demonstrate how the equation of a tangent to the circle at any point can be found without using calculus use the fact that the radius drawn from the point to the center is perpendicular to the tangent. Taking an appropriate point on the circle (2 , `sqrt 12)` , the slope of the radius from this point is equal to: `(sqrt 12 - 0)/(2 - 0)` = `sqrt 3`
The product of the slope of this line with the slope of a line perpendicular to this line is -1.
The slope of the tangent through (2, `sqrt 12)` is `-1/sqrt 3` . This gives the equation of the tangent as `(y - sqrt 12)/(x - 2) = -1/sqrt 3`
=> `sqrt3(y - sqrt 12) = 2 - x`
=> `x + sqrt3*y - 8 = 0`
This shows how you can find the equation of a tangent to any point on a circle without using calculus.
The circle defined by equation x^2 + y^2 =16 is one with center at origin (0,0).
This circle has a radius = sqrt(16) = 4
Point (2,2) lies on the circle with radius = sqrt(2^2 + 2^2)
This circle with radius sqrt(8) lies inside the given circle having radius sqrt(16), i.e. x^2 + y^2 = 16
Hence point (2,2) lies inside the given circle x^2 + y^2 = 16
Tangent cannot be drawn from point (2,2) to the circle.
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