JEE Advance - Mathematics (2020 - Paper 2 Offline - No. 3)
Let O be the centre of the circle x2 + y2 = r2, where $$r > {{\sqrt 5 } \over 2}$$. Suppose PQ is a chord of this circle and the equation of the line passing through P and Q is 2x + 4y = 5. If the centre of the circumcircle of the triangle OPQ lies on the line x + 2y = 4, then the value of r is .............
Answer
2
Explanation
As we know that the equation of family of circles passes through the points of intersection of given circle x2 + y2 = r2 and line PQ : 2x + 4y = 5 is,
(x2 + y2 $$-$$ r2) + $$\lambda $$(2x + 4y $$-$$ 5) = 0 ......(i)
Since, the circle (i) passes through the centre of circle
x2 + y2 = r2,
So, $$-$$ r2 $$-$$ 5$$\lambda $$ = 0
or 5$$\lambda $$ + r2 = 0 ....(ii)
and the centre of circle (i) lies on the line x + 2y = 4, so centre ($$-$$ $$\lambda $$, $$-$$ 2$$\lambda $$) satisfy the line x + 2y = 4.
Therefore, $$-$$$$\lambda $$ $$-$$4$$\lambda $$ = 4
$$ \Rightarrow $$ $$-$$5$$\lambda $$ = 4
$$ \Rightarrow $$ r2 = 4 {from Eq. (ii)}
$$ \Rightarrow $$ r = 2
(x2 + y2 $$-$$ r2) + $$\lambda $$(2x + 4y $$-$$ 5) = 0 ......(i)
Since, the circle (i) passes through the centre of circle
x2 + y2 = r2,
So, $$-$$ r2 $$-$$ 5$$\lambda $$ = 0
or 5$$\lambda $$ + r2 = 0 ....(ii)
and the centre of circle (i) lies on the line x + 2y = 4, so centre ($$-$$ $$\lambda $$, $$-$$ 2$$\lambda $$) satisfy the line x + 2y = 4.
Therefore, $$-$$$$\lambda $$ $$-$$4$$\lambda $$ = 4
$$ \Rightarrow $$ $$-$$5$$\lambda $$ = 4
$$ \Rightarrow $$ r2 = 4 {from Eq. (ii)}
$$ \Rightarrow $$ r = 2
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