Multiple choice

A circle with centre $(a, b)$ passes through the origin. The equation of the tangent to the circle at the origin is

  1. $ax - by = 0$
  2. $ax + by = 0$
  3. $bx- ay = 0$
  4. $bx + ay = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Circle center (a, b) passing through origin (0,0). Radius squared = a^2 + b^2. Equation: (x-a)^2 + (y-b)^2 = a^2 + b^2 => x^2 - 2ax + a^2 + y^2 - 2by + b^2 = a^2 + b^2 => x^2 + y^2 - 2ax - 2by = 0. Tangent at origin: set linear terms to 0, so -2ax - 2by = 0 => ax + by = 0.

AI explanation

The center of the circle is (a, b), so the slope of the radius connecting the center to the origin (0,0) is b/a. Because a tangent is perpendicular to the radius at the point of contact, the slope of the tangent line at the origin must be -a/b. Using the point-slope form with the point (0,0), the equation is y = (-a/b)x, which rearranges to ax + by = 0. The equation of the tangent is ax + by = 0.

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