Question

Prove that the line y = x + √2 a touches the circle x^2 + y^2 = a^2. Also find the point of contact.

Answer 1

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Answer 2

The line y = x + a√2 is tangent to circle x² + y² = a² and the point of contact is $(- \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}})$.

Step-by-step explanation:

If the line y = x + a√2 ............ (1)  is a tangent to the circle x² + y² = a² ............. (2), then by solving them we will get only one solution which will be the point of tangency.

Now, substituting the value of y from equation (1) to the equation (2) we get,

(x + a√2)² + x² = a²

⇒ x² + 2a² + 2(√2)ax + x² = a²

⇒ 2x² + 2(√2)ax + a² = 0

⇒ (x√2 + a)² = 0

⇒ $x = - \frac{a}{\sqrt{2} }$

Now, from equation (1) we get,

$y = - \frac{a}{\sqrt{2}} + a\sqrt{2} = \frac{a}{\sqrt{2}}$

Therefore, the solution is $(- \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}})$.

Therefore, the line (1) is tangent to the curve (2) and the point of contact is $(- \frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}})$. (Proved)