Question

If the equation of the tangent to the circle \(x^2+y^2=a^2\) at \(P(x_1, y_1)\) is \(x x_1 + y y_1 = K\), then K =

• A. \( a^2 \)
• B. \( r^2 \) (r is usually radius, so if a is radius, this is same as (a))
• C. \( a \)
• D. \( 2a^2 \)

Hint:

The equation of the tangent to the circle \(x^2+y^2=r^2\) at a point \((x_1, y_1)\) on the circle is \(xx_1 + yy_1 = r^2\).
For a general circle \(x^2+y^2+2gx+2fy+c=0\), the tangent at \((x_1,y_1)\) is \(xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\).
This rule involves replacements: \(x^2 \to xx_1\), \(y^2 \to yy_1\), \(2x \to x+x_1\), \(2y \to y+y_1\).

Answer 1

The Correct Option is A

The equation of the circle is \(x^2+y^2=a^2\). This circle is centered at the origin (0,0) and has radius \(r=a\). The point \(P(x_1, y_1)\) lies on the circle, so it satisfies the circle's equation: \(x_1^2 + y_1^2 = a^2\). The equation of the tangent to the circle \(x^2+y^2=a^2\) at a point \((x_1, y_1)\) on the circle is given by: \[ xx_1 + yy_1 = a^2 \] This is obtained by replacing \(x^2\) with \(xx_1\), \(y^2\) with \(yy_1\) in the circle equation. The question states that the equation of the tangent is \(x x_1 + y y_1 = K\). Comparing this with the standard equation of the tangent \(xx_1 + yy_1 = a^2\), we can see that: \[ K = a^2 \] Option (a) is \(a^2\). Option (b) is \(r^2\). Since the radius of the circle \(x^2+y^2=a^2\) is \(a\), then \(r=a\), so \(r^2=a^2\). Thus, both (a) and (b) represent the same value if \(r\) is taken as the radius of this specific circle. Assuming the options are distinct and the standard form uses \(a^2\), then \(K=a^2\). \[ \boxed{a^2} \]