Question

If the line y=mx+c is a tangent to the circle x2+y2=a2,then the point of contact is​

Answer 1

Answer:

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The equation of the line is

y=mx+c

⇒ mx−y+c=0 ....(i)

Here, a=m,b=−1,c=c

The equation of the circle is

x2+y2=a2 ...(ii)

Line (i) will be a tangent to the circle (ii), if the length of perpendicular from the centre (0,0) of circle (ii) on line (i) = the radius a.

⇒±m2+1m(0)−0+c=a

⇒c=±a(1+m2)

⇒c2=a2(1+m2)

Which is the required condition.

Answer 2

Given: The line y=mx+c is a tangent to the circle x2+y2=a2.

To find: The point of contact

Solution: Let (x₁ , y₁) be the point of contact of the tangent y = mx + c with the circle x² + y² = a².

The let us consider y₁ = mx₁ + c to be equation i.

Now, the equation of the tangent at the point (x₁ , y₁) is xx₁ + yy₁ = a².

So yy₁ = -xx₁ + a² [equation ii].

The equations i and ii represent the same line and hence their coefficients are proportional.

Hence, y₁/1 = -x₁/m = a²/c

        ⇒ y₁ = a²/c,

            x₁ = - a²m/c,

            c = ± a√(1 + m²).

Then, the point of contact is either

(-am/√(1 + m²) , a/√(1 + m²)) or (am/√(1 + m²), -a/√(1 + m²)).

Answer: (-am/√(1 + m²) , a/√(1 + m²)) or (am/√(1 + m²), -a/√(1 + m²))