If the circles x^2 + y^2 + 2ax + cy + a = 0 a | vedclass

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(D) The equation of the common chord $PQ$ is obtained by subtracting the two circle equations: $(x^2 + y^2 + 2ax + cy + a) - (x^2 + y^2 - 3ax + dy - 1

Vedclass · Accepted Answer

No value of $a$

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(D) The equation of the common chord $PQ$ is obtained by subtracting the two circle equations: $(x^2 + y^2 + 2ax + cy + a) - (x^2 + y^2 - 3ax + dy - 1) = 0$.This simplifies to $5ax + (c - d)y + (a + 1) = 0$.....$(i)$The given equation of the line passing through $P$ and $Q$ is $5x + by - a = 0$.....$(ii)$Since both equations represent the same line,their coefficients must be proportional:$\frac{5a}{5} = \frac{c - d}{b} = \frac{a + 1}{-a}$From the first and third parts: $a = \frac{a + 1}{-a}$$-a^2 = a + 1$$a^2 + a + 1 = 0$For the quadratic equation $a^2 + a + 1 = 0$,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(1) = -3$.Since $D < 0$,there are no real values of $a$ that satisfy this condition.

If the circles $x^2 + y^2 + 2ax + cy + a = 0$ and $x^2 + y^2 - 3ax + dy - 1 = 0$ intersect in two distinct points $P$ and $Q$,then the line $5x + by - a = 0$ passes through $P$ and $Q$ for

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