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

Circle PYQ — JEE Main 2005 · Shift 0 · Q105 with Solution

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Correct answer: B

Given circles

The two circles are $S_1:x^2+y^2+2ax+cy+a=0$ and $S_2:x^2+y^2-3ax+dy-1=0.$

They intersect at two distinct points $P$ and $Q$ .

The common chord $PQ$ is given by the radical axis, obtained by subtracting the two equations.

Equation of the common chord $PQ$

Compute $S_1-S_2=0$ : $\left(x^2+y^2+2ax+cy+a\right)-\left(x^2+y^2-3ax+dy-1\right)=0$

Simplifying, $2ax+3ax+cy-dy+a+1=0$ $5ax+(c-d)y+(a+1)=0.$

So the line through $P$ and $Q$ is $5ax+(c-d)y+(a+1)=0.$

Given line

The line $5x+by-a=0$ passes through both $P$ and $Q$ .

Since $P,Q$ are the two intersection points of the circles, the line through them must be the same as the radical axis. Hence $5x+by-a=0$ must be proportional to $5ax+(c-d)y+(a+1)=0.$

So there exists some nonzero constant $\lambda$ such that

\quad c-d=b\lambda, \quad a+1=-a\lambda.$$ From the first relation, $$\lambda=a.$$ Substitute into the third relation: $$a+1=-a\cdot a=-a^2.$$ Thus $$a^2+a+1=0.$$ --- 4. **Check for real values of $a$** The quadratic equation is $$a^2+a+1=0.$$ Its discriminant is $$\Delta=1-4=-3<0.$$ Hence there is **no real value** of $a$ satisfying this condition. So the line $5x+by-a=0$ cannot pass through both intersection points $P,Q$ for any real $a$. --- 5. **Conclusion** Therefore, the correct option is $$\boxed{\text{B: no value of }a}.$$ --- 6. **Comparison with stored answer** Stored correct answer: **B** This matches our derived answer.

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View written solutionFree

Correct answer: B

Given circles

The two circles are $S_1:x^2+y^2+2ax+cy+a=0$ and $S_2:x^2+y^2-3ax+dy-1=0.$

They intersect at two distinct points $P$ and $Q$ .

The common chord $PQ$ is given by the radical axis, obtained by subtracting the two equations.

Equation of the common chord $PQ$

Compute $S_1-S_2=0$ : $\left(x^2+y^2+2ax+cy+a\right)-\left(x^2+y^2-3ax+dy-1\right)=0$

Simplifying, $2ax+3ax+cy-dy+a+1=0$ $5ax+(c-d)y+(a+1)=0.$

So the line through $P$ and $Q$ is $5ax+(c-d)y+(a+1)=0.$

Given line

The line $5x+by-a=0$ passes through both $P$ and $Q$ .

Since $P,Q$ are the two intersection points of the circles, the line through them must be the same as the radical axis. Hence $5x+by-a=0$ must be proportional to $5ax+(c-d)y+(a+1)=0.$

So there exists some nonzero constant $\lambda$ such that

\quad c-d=b\lambda, \quad a+1=-a\lambda.$$ From the first relation, $$\lambda=a.$$ Substitute into the third relation: $$a+1=-a\cdot a=-a^2.$$ Thus $$a^2+a+1=0.$$ --- 4. **Check for real values of $a$** The quadratic equation is $$a^2+a+1=0.$$ Its discriminant is $$\Delta=1-4=-3<0.$$ Hence there is **no real value** of $a$ satisfying this condition. So the line $5x+by-a=0$ cannot pass through both intersection points $P,Q$ for any real $a$. --- 5. **Conclusion** Therefore, the correct option is $$\boxed{\text{B: no value of }a}.$$ --- 6. **Comparison with stored answer** Stored correct answer: **B** This matches our derived answer.

Circle question

Circle question

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Circle question

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