Question 37

Let the circle $$x^2 + y^2 = 2ax + 2by$$ intersect the $$x$$-axis at point $$A(\alpha, 0)$$ and $$y$$-axis at point $$B(0, \beta)$$, where $$\alpha\beta \neq 0$$. If the point $$C(p, q)$$ lies on the chord AB, then $$\frac{p+\alpha}{a} + \frac{q+\beta}{b}$$ equals

Solution

Setting $$y=0$$: $$x^2 - 2ax = 0 \Rightarrow x \in \{0, 2a\}$$, so $$\alpha = 2a$$.

Setting $$x=0$$: $$\beta = 2b$$.

Line $$AB$$: $$\dfrac{x}{2a} + \dfrac{y}{2b} = 1$$. Point $$C(p,q)$$ on chord: $$\dfrac{p}{2a} + \dfrac{q}{2b} = 1$$.

$$\dfrac{p + \alpha}{a} + \dfrac{q + \beta}{b} = \dfrac{p}{a} + 2 + \dfrac{q}{b} + 2 = 2\left(\dfrac{p}{2a} + \dfrac{q}{2b}\right) + 4 = 2 + 4 = \mathbf{6}$$.

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Let the circle $$x^2 + y^2 = 2ax + 2by$$ intersect the $$x$$-axis at point $$A(\alpha, 0)$$ and $$y$$-axis at point $$B(0, \beta)$$, where $$\alpha\beta \neq 0$$. If the point $$C(p, q)$$ lies on the chord AB, then $$\frac{p+\alpha}{a} + \frac{q+\beta}{b}$$ equals

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