If a line, y â mx + c is a tangent to the circle, (x â 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2+y2=1 at the point ((1/√2), (1/√2)) ; then :

Question

If a line, y â mx + c is a tangent to the circle, (x â 3)2 + y2 = 1 and it is perpendicular to a line L1, where L1 is the tangent to the circle, x2+y2=1 at the point ((1/√2), (1/√2)) ; then : (A) c2+7c + 6 = 0 (B) c2-6c + 7 = 0 (C) c2-7c + 6 = 0 (D) c2+6c + 7 = 0

Tardigrade Admin · Accepted Answer

Slope of tangent to x2 + y2 = 1 at (P/ )((1/√2), (1/√2)) 2x + 2yy' = 0 ⇒ mT P = -1 y = mx + c is tangent to (x - 3)2 + y2 = 1 y = x + c is tangent to (x - 3)2 + y2 = 1 |(c+3/√2)| = 1 ⇒ c2 + 6c + 7 = 0

Q. If a line, $y — m x + c$ is a tangent to the circle, $(x —3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2});$ then :

$c^{2} + 7 c + 6 = 0$

$c^{2} - 6 c + 7 = 0$

$c^{2} - 7 c + 6 = 0$

$c^{2} + 6 c + 7 = 0$

Slope of tangent to $x^{2} + y^{2} = 1$ at $P (\frac{1}{2}, \frac{1}{2})$
$2 x + 2 y y^{'} = 0 \Rightarrow m_{TP} = - 1$
$y = m x + c$ is tangent to $(x - 3)^{2} + y^{2} = 1$
$y = x + c$ is tangent to $(x - 3)^{2} + y^{2} = 1$
$∣ ∣ \frac{c + 3}{2} ∣ ∣ = 1 \Rightarrow c^{2} + 6 c + 7 = 0$

Q. If a line, y—mx+c is a tangent to the circle, (x—3)2+y2=1 and it is perpendicular to a line L1​, where L1​ is the tangent to the circle, x2+y2=1 at the point (2​1​,2​1​); then :

c2+7c+6=0

c2−6c+7=0

c2−7c+6=0

c2+6c+7=0