Question

A variable circle C has the equation $x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + t = 0$, where t is a parameter. If the power of point P(a,b) w.r.t. the circle C is constant then the ordered pair (a, b) is

• A. $(\frac{1}{10}, \frac{1}{10})$
• B. $(-\frac{1}{10}, \frac{1}{10})$
• C. $(\frac{1}{10}, \frac{1}{10})$
• D. $(-\frac{1}{10}, \frac{1}{10})$

Answer 1

Answer: (B), (D)

$(-\frac{1}{10}, \frac{1}{10})$

Answer 2

The power of a point $P(a,b)$ with respect to a circle $x^2 + y^2 - 2gx - 2fy + c = 0$ is given by $a^2 + b^2 - 2ga - 2fb + c$. For the given circle $x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + t = 0$, we have $g = t^2 - 3t + 1$, $f = t^2 + 2t$, and $c = t$. The power of $P(a,b)$ is $a^2 + b^2 - 2(t^2 - 3t + 1)a - 2(t^2 + 2t)b + t$. Expanding and grouping terms by powers of $t$: Power$(P) = a^2 + b^2 - 2at^2 + 6at - 2a - 2bt^2 - 4bt + t$ Power$(P) = (-2a - 2b)t^2 + (6a - 4b + 1)t + (a^2 + b^2 - 2a)$ For the power to be constant with respect to $t$, the coefficients of $t^2$ and $t$ must be zero.

1. Coefficient of $t^2$: $-2a - 2b = 0 \implies a + b = 0 \implies b = -a$.
2. Coefficient of $t$: $6a - 4b + 1 = 0$. Substituting $b = -a$ into the second equation: $6a - 4(-a) + 1 = 0 \implies 6a + 4a + 1 = 0 \implies 10a + 1 = 0 \implies a = -\frac{1}{10}$. Since $b = -a$, $b = -(-\frac{1}{10}) = \frac{1}{10}$. Thus, the ordered pair $(a,b)$ is $(-\frac{1}{10}, \frac{1}{10})$.