Question

From a point P outside of a circle with centre at C tangents PX and PY are drawn such that $\frac { 1 } { ( \mathrm { CX } ) ^ { 2 } }$+$\frac { 1 } { ( \mathrm { PX } ) ^ { 2 } }$=$\frac { 1 } { 16 }$then the length of chord XY is –

• A. 8
• B. 12
• C. 16
• D. None of these

Answer 1

Answer: (A)

8

Answer 2

Let PX = t, XC = r, PC = c and XM = k

Now in DPXC and DXMC

$\frac { \mathrm { t } } { \mathrm { c } }$ = $\frac { \mathrm { k } } { \mathrm { r } }$Ž tr = ck

Now given $\frac { 1 } { \mathrm { r } ^ { 2 } }$ + $\frac { 1 } { \mathfrak { t } ^ { 2 } }$ = $\frac { 1 } { 16 }$and using k =

Ž k² = = $\frac { 1 } { \frac { 1 } { \mathrm { t } ^ { 2 } } + \frac { 1 } { \mathrm { r } ^ { 2 } } }$ = 16

Ž k = 4

So xy = 2k = 8