Equal circles with centres O and O' touch each other at X. OO' produced to meet a circle with centre O , at A. AC is a tangent to the circle whose centre is O. O'D is perpendicular to AC. Find the value of DO'/CO.....cAn aNy1 hElP MeH WiD thIs qUeStIoN pLs...???????
Hi parvathi ! :)
Here's how we did it. Hope you find it fine ! :D
And for the diagram of the question , check pg 10.31 of R.D Sharma.
Given : Two circles with centres O and O' are equal , which means that their respective radii are eual to each other. So , XO = XO'
2. Angle ADO'= 90
Proof : Since that tangent passes through the circle with centre O at C , Angle OCD = 90 ( Radius perpendicular to tangent )
Now , we look at the two triangles in the figure , Triangle ADO' and triangle AOC.
In ADO' ---> AO' is the hypotenuse , as it is the side opp to the right angle.
Hence using the pythagoras theorem we can say that - AO' square = DO' square + AD square
So , DO' = root of (AD square - AO'square)
In Triangle AOC ---> By the pythagoras theorem , we can say that - root of (AO square - ACsquare )= CO
Now we can take AO' as 'x' to make the steps ahead easier.
If AO' = x
then AO = 3x ( because the radii are equal , so AO = AO'+O'X+OX)
In the previous classes , we have learnt that "a line parallel to a side of a triangle intersects the other side in the same ratio".(pg 124 of tb)
Hence we can say that since A0'/O'D = x/2x = 1/2 ---> AD/DC= 1/2 and DC/2 = AD
Now , we substitute these values into the two pythagoras equations..
So , root of ( X square - DC square/4) = OD -- 1.
And 9x square - (DC/2+ DC)square = OC square
-->root of ( 9x square - 9DC square/4 ) = OC --- 2.
Dividing 1. by 2. , we get
Root of ( 4x square - DC square ) / Root of (36 DC square - 9DC square)
= Root of (I/9)
@parvathy_meera. such an elaborate answer is not needed . @Parvathi Sudeephope this helps you !
@Nikiii Nikhilhope this is easier to understand !
Please check pg no.10.31 in RD Sharma (Qn.29) for the diagram.
angleADO'=90 ( since O'D is perpendicular to AC) angleACO= 90 ( OC(radius)perpendicular to AC(tangent))
In triangles ADO'and ACO ,
angleADO'=angleACO ( each 90)
angle DAO = angle CAO (common)
by AA criterion ,triangles ADO' and ACO are similar to each other
AO'/AO=DO'/CO ( corresponding parts of similar triangles )
=3AO'(since AO'=O'X=OX because radii of the two circles are equal )