GATE Exam - GATE Preparation - GATE IIT Test Series - Online Engineering Expert - GATE Forum - GATE Syllabus

Transmigrator

Two circles are drawn through the points (a, 5a) and (4a, a) to touch the axis of y. Prove that they intersect at an angle of tan inverse 40/5.

It would be kind of u to solve this............................

Transmigrator

Plz solve.................Rates assured

Anant Kumar

Let $A$ be the point $(a,\,5a)$ and $B$ be $(4a,\,a)$ . Any circle that touches the $y$ -axis will have the $x$ -coordinate of its center same (in absolute value) as the radius.

The mid point of $AB$ is $C_1(5a/2,\,3a)$ while the length of $AB=5a$ so that half of $AB$ is $5a/2$ which is same as the $x$ -coordinate. So one of the circles passing through $A$ and $B$ and touching the $y$ axis will have $AB$ as its diameter. Its center is $C_1$ and radius $r_1=5a/2$ .

Let the other circle be centered at $C_2(h,\,k)$ . Then, $C_2A=C_2B$ , which gives

$(h-a)^2 + (k-5a)^2 = h^2$ and $(h-4a)^2 + (k-a)^2 = h^2$

From these we obtain

$3k^2 -38ak -87a^2=0$

Sum of roots $= 38a/3$

Obviously, the $y$ coordinate of the center of the first circle is a root of this quadratic (it can be checked by direct substitution), we get the ordinate of $C_2$ as $k=\dfrac{38}{3}- 3a = \dfrac{29a}{3}$ . So that $h=\dfrac{205a}{18}$ . Thus, $r_2=\dfrac{205a}{18}$

Now the angle between the tangents will be same as the angle between the corresponding radii i.e. between $C_1A$ and $C_2A$ . We see that if this angle be $\theta$ , then

$\cos\theta = \dfrac{C_1A}{C_2A}=\dfrac{r_1}{r_2}=\dfrac{5a/2}{205a/18}=\dfrac{5\times 18}{2\times 205}=\dfrac{9}{41}$

Therefore, $\tan\theta = \sqrt{\dfrac{1}{\cos^2\theta}-1}=\sqrt{\dfrac{41^2}{9^2}-1}=\sqrt{\dfrac{1681-81}{9^2}}=\dfrac{40}{9}$

which was to be proved.

Anant Kumar

The question has been posted wrongly. Its 40/9 and not 40/5.

Transmigrator

Thanx Sir............This problem was really irritating me

Ask & Discuss Questions with Community & Experts