If the centre of a circle is (-6, 8) and it passes through the origin, then equation to its tangent at the origin isa)2y = xb)4y = 3xc)3y = 4xd)3x + 4y = 0Correct answer is option 'B'. Can you explain this answer?

Question

Answer

Solution:

Given, the centre of a circle is (-6, 8) and it passes through the origin.

Let radius of the circle be 'r'.

As the circle passes through the origin, therefore the distance between the centre and the origin is equal to the radius of the circle.

So, we have

√[(-6 - 0)² + (8 - 0)²] = r

r = √100 = 10

Hence, the equation of the circle is (x + 6)² + (y - 8)² = 100.

Differentiating the above equation with respect to x, we get

2(x + 6) + 2(y - 8)dy/dx = 0

At the point (0,0), we have

2(0 + 6) + 2(0 - 8)dy/dx = 0

dy/dx = 3/4

Therefore, the equation of the tangent at the origin is

y - 0 = (3/4)(x - 0)

4y = 3x

Hence, option (B) is the correct answer.

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