There are two circles with centers A and B having radii 5 cm and 3 cm.

Let O be the point on AB that bisects AB. AP and AQ would be the radius of the larger circle which is 5.
AB is the difference between the radius of the larger circle and the smaller circle which is 5-3= 2
Thus we have two right triangles with hypotenuse as 5 and one leg as 1.
Thus OP^2= OQ^2= 5^2- 1^2
OP= OQ= square root of 24
Thus, PQ= 2* square root of 24
PQ= 4√6

Answer: C

given are two circles with centers A and B having radii 5 cm and 3 cm touching internally
The perpendicular bisector of segment AB meets the bigger circle at P and Q. Target Find the length of PQ.
Let the point of touch of two circles be C ; so AC = 5cm and BC = 3cm
also a LINE PQ which for circle center A pass through the line AC and intersect it at point D
we get AC-BC = AB ; 5-3 ; 2 cm
and PQ is perpendicular bisector so AD = 1 cm
now for ∆ PAD ; PA = 5cm and AD = 1 cm we can determine PD ; i.e 25-1 ; 24 ; 2√6
and since PQ is a chord of circle so its distance will be 2 *PD ; 2*2√6 ; 4√6
OPTION C


There are two circles with centers A and B having radii 5 cm and 3 cm. They touch each other internally. If the perpendicular bisector of segment AB meets the bigger circle at P and Q. Find the length of PQ.

A. 2√3
B. 3√4
C. 4√6
D. 5√6
E. 5√7

There are two circles with centers A and B having radii 5 cm and 3 cm. They touch each other internally. If the perpendicular bisector of segment AB meets the bigger circle at P and Q. Find the length of PQ.

A. 2√3
B. 3√4
C. 4√6
D. 5√6
E. 5√7

Solution:



OAP is a right-angled triangle, in which, OA = 1 and AP = 5,

Therefore, \(OP^2 = AP^2 - AO^2 = 25 - 1 ; OP = \sqrt{24} = 2\sqrt{6}\)
PQ = 2 OP = \(4\sqrt{6}\)

So, C is the correct answer

length of AB = 2cm

perpendicular bisector of AB divides AB in two equal parts at point O.
in one of triangles formed by perpendicular bisector,
AP^2 = AO^2 + OP^2
OP^2 = 25 - 1 = 24; OP = 2√6

similarly, in other triangle, OQ also = 2√6

PQ = OP + OQ = 4√6
Ans: C

Let line segment AB and PQ cut each other at M
Let AM = X, BM = Y, PM = QM = Z
In triangle APM,
AP^2 = AM^2+PM^2
5^2 = X^2+Z^2
25-X^2 = Z^2 -------(1)

BP^2 = BM^2+PM^2
3^2 = Y^2+Z^2
9-Y^2 = Z^2 -------(2)

25-X^2 = 9-Y^2
16 = X^2-Y^2 --------(3)

A. 2√3 = 2Z => Z = √3
Putting value of Z in equation 1&2 respectively
25-3 = 22 = X^2 &
9-3 = 6 = Y^2.
Again putting value of X^2 & Y^2 in equation 3
22-6 = 16 => 16 = 16 (matching)

B. 3√4 = 2Z => Z = 3
Putting value of Z in equation 1&2 respectively
25-9 = 16 = X^2 &
9-9 = 0 = (zero value, which cannot be null)

C. 4√6 = 2Z => Z = 2√6
Putting value of Z in equation 1&2 respectively
25-12 = 13 = X^2 &
9-12 = (negative value, which cannot be negative)

D. 5√6 = 2Z => Z = 5√6/2
Putting value of Z in equation 1&2 respectively
25-75 = (negative value, which cannot be negative)

E. 5√7 = 2Z => Z = 5√7/2
Putting value of Z in equation 1&2 respectively
25-87.5 = (negative value, which cannot be negative)

IMO A

Posted from my mobile device

OA is 4root6 . again pythagoras theorem is to be used

Given:
1. There are two circles with centers A and B having radii 5 cm and 3 cm.
2. They touch each other internally.
3. The perpendicular bisector of segment AB meets the bigger circle at P and Q.

Asked: Find the length of PQ.



AB = 5 - 3 = 2
\(AT = TB = \frac{AB}{2} = 1\)

\(In \triangle PTB\)

\(\angle PTB = 90^0\)
TB = 1
PB = radius of bigger circle = 5
\(PT = \sqrt{PB^2 - TB^2} = \sqrt{5^2 - 1^2} = \sqrt{24} = 2\sqrt{6}\)

\(PQ = PT*2 = 4\sqrt{6}\)

IMO C

c,IMO

AB = 2 . So PQ/2 = root (24). so PQ = 2 root(24) = 2 * 2 (root (6)) = 4 root(6).
So c .

Posted from Mobile device

Let the meeting point of two circles be Y.

AY=AB+BY..........(AY=Radii of the bigger circle 5 cm. BY=Radii of the smaller circle 3cm)
\(\therefore AB = 5-3 = 2cm.\)

A perpendicular bisector of segment AB will divide AB & itself into two equal parts...(AX=XB, PX=XQ) Let the perpendicular bisector meet AB at X.
\(\therefore AX=XB=1cm\)

In \\(triangle APX\), AP = Radii of bigger circle 5cm, AX=1cm.
By Pythagoras theorem, \(PX=2\sqrt 6\)

PQ = PX+XQ
PQ = 2PX