Question

The squared length of the intercept made by the line x = h on the pair of tangents drawn from the origin to the circle

x² + y² + 2gx + 2fy + c = 0 is

• A. $\frac { 4 \mathrm { ch } ^ { 2 } } { \left( \mathrm {~g} ^ { 2 } - \mathrm { c } \right) ^ { 2 } }$ (g² + f² – c)
• B. $\frac { 4 \mathrm { ch } ^ { 2 } } { \left( \mathrm { f } ^ { 2 } - \mathrm { c } \right) ^ { 2 } }$ (g² + f² – c)
• C. $\frac { 4 \operatorname { ch } ^ { 2 } } { \left( \mathrm {~g} ^ { 2 } - \mathrm { f } ^ { 2 } \right) ^ { 2 } }$ (g² + f² – c)
• D. None of these

Answer 1

Answer: (B)

$\frac { 4 \mathrm { ch } ^ { 2 } } { \left( \mathrm { f } ^ { 2 } - \mathrm { c } \right) ^ { 2 } }$ (g² + f² – c)

Answer 2

Equation of pair of tangents from (0, 0) is

SS¢ = T² Ž (x² + y² + 2gx + 2fy + c)

c = (gx + fy + c)² ... (i)

The intersection points of the above pair with x = h is given by

(gh + fy + c)² = (h² + y² + 2gh + 2fy + c) c

Ž (f² – c) y² + 2fghy + h² (g² – c) = 0

If its roots are y₁ and y₂, then length of intercept

AB² = |y₁ – y₂|² = (y₁ + y₂)² – 4y₁y₂

= $\left( \frac { 2 \mathrm { fgh } } { \mathrm { f } ^ { 2 } - \mathrm { c } } \right) ^ { 2 }$ – 4 = (g² + f² – c)