The circle above has center X and radius 10. If PB = 8 and QD = 6, wha

(Draw a line from X to D = XD. Which will ne radius)
In triangle XQD
XQ^2= XD^2-QD^2
XQ^2= 10^2-6^2
XQ^2= 100-36
XQ^2=64
XQ= 8

Since XR is also Radius.
XR = XQ+ QR
10= 8+QR
2= QR

Answer is 2

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B.
Join DX.
In triangle QDX; angle QDX = 90
Using Pythagoras theorem: QX^2 + QD^2 = XD^2
XD = radius = 10
QX^2 + 6^2 = 10^2
Solving:: QX = 8

NOW: XR = RQ + QX
XR = radius =10
10 = 8 + RQ
RQ or QR = 2
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We know that XR is the radius, so to find QR we need to distance XQ as \(QR = XR - XQ\)

Let's take triangle XQD. Joining X and D we get XD as the radius that is equal to 10.
XQD is a right angled triangle so applying pythagoras theorem.

\(XQ^2 + QD^2 = XD^2\)
\(XQ^2 + 6^2 = 10^2\)
\(XQ^2 + 36 = 100\)
\(XQ^2 = 100 - 36\)
\(XQ^2 = 64\)
\(XQ = 8\)

\(QR = XR - XQ\)
\(QR = 10 - 8\)

\(QR = 2\)

IMO, B

as XB is drawn, draw XD

XB=XD=XR radius of circle =10 ---- i

let RQ =x ----- ii

Now, XR= XQ+QR
10 = XQ+x
or XQ= 10-x ------- iii

Triangle XQD is a right angle triangle and hence
XD^2 = XQ^2 + QD^2
putting value of XD from i, and QD= 6 as given
we get
10^2 = (10-x)^2 + 6^2
or
100-36= 64= 8^2 = (10-x)^2
or
10-x=8 or x=2 = RQ

Option B,

We join XD, which becomes Radius.
XD = 10.
QD = 6 ( Given ).
Using Pythagorean triplets, QX = 8.
RX = 10 ( Radius ).
QR = RX - QX.
QR = 10 - 8 = 2 ( Ans ).

GIVEN:

• A circle with center X and radius 10.
• PB = 8
• QD = 6.

ASKED:

• Length of QR.

APPROACH:


Observe in the given figure that R is a point on the circle of center X. This means that XR is a radius, and hence, XR = 10.
For an instant, let’s just focus on the radius XR that contains our needed QR.



We can think of finding QR in two ways:
- Directly, or
- by finding XQ and using QR = XR – XQ …. (1)

No direct means of finding QR exists. Even if we look at the triangle RQD, we only have QD given. With two unknown sides, we cannot find QR.
Instead, we can find the length of XQ from the triangle QXD using Pythagoras’ theorem since we know the lengths of two sides QD and XD. Then, XR – XQ will give us the length of QR.

WORKING OUT:
Just as we discussed in the approach, we are going to find the length of XQ first and then use the formula QR = XR – XQ.
Since we already know XR = 10, we just need to find XQ.

Finding XQ:
Join points X and D to construct triangle QXD (drawn below).
Then, since D is a point on the circle whose center is X, XD is also a radius. So, XD = 10. (As marked in triangle QXD below)


Since we now have two sides of the right triangle QXD, we can use Pythagoras’ Theorem to get XD:

• XQ2 + QD2 = XD2
• XQ2 + 62 = 102
• XQ2 = 100 – 36 = 64

XQ = 8 (Since -8 is not a possible length)

Final Answer:
Using the value of XQ = 8 in eq (1), QR = XR – XQ, we get:

• QR = 10 – 8
• QR = 2

Correct Choice: B

Best Regards,
Ashish Arora
Quant Expert, e-GMAT
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