Question

Chord AB of the circle $x^2 + y^2 = 100$ passes through the point (7, 1) and subtends an angle of $60^\circ$ at the circumference of the circle. If $m_1$ and $m_2$ are the slopes of two such chords then the value of $m_1m_2$ is

• A. -1
• B. 1
• C. 7/12
• D. -3

Answer 1

Answer: (A)

-1

Answer 2

The equation of the circle is $x^2 + y^2 = 100$, with center $O(0, 0)$ and radius $R = 10$. The chord AB passes through the point $P(7, 1)$. The angle subtended by the chord AB at the circumference is $60^\circ$, which means the angle subtended at the center is $2 \times 60^\circ = 120^\circ$. The perpendicular distance $d$ from the center to the chord can be found using trigonometry in the isosceles triangle formed by the center and the chord endpoints. In a right-angled triangle formed by the center, the midpoint of the chord, and one endpoint of the chord, the angle at the center is $120^\circ/2 = 60^\circ$. Thus, $d = R \cos(60^\circ) = 10 \times \frac{1}{2} = 5$.

Let the equation of a chord passing through $P(7, 1)$ with slope $m$ be $y - 1 = m(x - 7)$, which can be rewritten as $mx - y + (1 - 7m) = 0$. The distance from the center $O(0, 0)$ to this line is given by the formula $d = \frac{|m(0) - (0) + (1 - 7m)|}{\sqrt{m^2 + (-1)^2}} = \frac{|1 - 7m|}{\sqrt{m^2 + 1}}$. Since $d=5$, we have $5 = \frac{|1 - 7m|}{\sqrt{m^2 + 1}}$. Squaring both sides gives $25(m^2 + 1) = (1 - 7m)^2$. Expanding and rearranging, we get $25m^2 + 25 = 1 - 14m + 49m^2$, which simplifies to $24m^2 - 14m - 24 = 0$. Dividing by 2, we obtain the quadratic equation $12m^2 - 7m - 12 = 0$. The roots of this equation, $m_1$ and $m_2$, are the slopes of the two possible chords. For a quadratic equation $am^2 + bm + c = 0$, the product of the roots is $c/a$. In this case, $a=12$, $b=-7$, and $c=-12$. Therefore, the product of the slopes $m_1m_2 = \frac{-12}{12} = -1$.