Let the circle x^2 + y^2 = 2ax + 2by intersect the x-axis at point A(, 0) and y-axis at point B(0, ), where ≠ 0. If the point C(p, q) lies on the chord AB, then (p+)/a + (q+)/b equals Option 1: 4 Option 2: 2 Option 3: 6 Option 4: 3

Answer: 6. Solution: Rewrite the circle as x^2 + y^2 - 2ax - 2by = 0. Set y = 0: x^2 - 2ax = 0, giving x = 0 or x = 2a. So = 2a. Set x = 0: y^2 - 2by = 0, giving y = 0 or y = 2b. So = 2b. Line AB joins (2a, 0) and (0, 2b). Its intercept form is: x/2a + y/2b = 1 Multiply by 2: x/a + y/b = 2 Since C(p, q) lies on AB: p/a + q/b = 2 Now compute the required expression: (p + )/a + (q + )/b = p/a + 2a/a + q/b + 2b/b = (p/a + q/b) + 2 + 2 = 2 + 4 = 6 Answer = 6

Geometry

Circles

Hard

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

✅ Correct Option: 3

Rewrite the circle as $x^2 + y^2 - 2ax - 2by = 0$ .

Set $y = 0$ : $x^2 - 2ax = 0$ , giving $x = 0$ or $x = 2a$ . So $\alpha = 2a$ .

Set $x = 0$ : $y^2 - 2by = 0$ , giving $y = 0$ or $y = 2b$ . So $\beta = 2b$ .

Line $AB$ joins $(2a, 0)$ and $(0, 2b)$ . Its intercept form is:

$\dfrac{x}{2a} + \dfrac{y}{2b} = 1$

Multiply by $2$ :

$\dfrac{x}{a} + \dfrac{y}{b} = 2$

Since $C(p, q)$ lies on $AB$ :

$\dfrac{p}{a} + \dfrac{q}{b} = 2$

Now compute the required expression:

$\dfrac{p + \alpha}{a} + \dfrac{q + \beta}{b} = \dfrac{p}{a} + \dfrac{2a}{a} + \dfrac{q}{b} + \dfrac{2b}{b}$

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

Answer $= 6$

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