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

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(B) 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

For no value of $a$

Vedclass · Answer

(B) 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 $5ax + (c - d)y + (a + 1) = 0$ ...... $(i)$ We are given that the line $PQ$ is $5x + 6y - a = 0$ ...... (ii) Comparing the coefficients of $(i)$ and (ii),we have: $\frac{5a}{5} = \frac{c - d}{6} = \frac{a + 1}{-a}$. From the first and third parts: $a = \frac{a + 1}{-a} \implies -a^2 = a + 1 \implies a^2 + a + 1 = 0$. The discriminant of this quadratic equation is $D = 1^2 - 4(1)(1) = -3$. 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 at two distinct points $P$ and $Q$,then for what value of $a$ does the line $5x + 6y - a = 0$ pass through points $P$ and $Q$?

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