Question

The tangent to the circle \(\rm x^2+y^2=16\), which is parallel to y-axis and does not lie in first quadrant, touches the circle at the point

1. (-4, 0)
2. (4, 0)
3. (0, -4)
4. (0, 4)

Answer 1

Correct Answer - Option 1 : (-4, 0)

Concept:

The equation of a straight line parallel to y-axis at a distance 'a', is x = a

Perpendicular Distance of a Point from a Line

Let us consider a line given by, Ax + By + C = 0

And a point whose coordinate is (x₁, y₁)

Now, distance =\( \rm d=|\frac{Ax_1+By_1+c}{\sqrt{A^2+B^2}}| \)

 

Calculation:

Any line parallel to y-axis is x = a

If it touches the circle \(\rm x^2+y^2=16\), the perpendicular distance from the center (0,0) of the circle to the line x = a⇒ x - a = 0, must be equal to radius 4.

∴ \( \rm 4=|\frac{0-a}{1}|\)

\(\)⇒  a = ±4

 Tangent does not lie in first quadrant

∴ a = -4

∴ Equation of tangent is x = -4

It touches the circle when \(\rm 16+y^2=16\Rightarrow y =0\)

∴ It touches the circle at the point (-4, 0)

Hence, option (1) is correct.