A common tangent of the two circles is, A tangent is drawn to the circle at the point . A straight line , perpendicular to , is a tangent to the circle . [2012]

Question

A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ , is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A common tangent of the two circles is

Smart Achievers · Accepted Answer

(4)

From the figure it is clear that the intersection point of two direct common tangents lies on the $x$ -axis.

$Also ∆ P T_{1} C_{1} ~ ∆ P T_{2} C_{2} \Rightarrow P C_{1} : P C_{2} = 2 : 1$

$Hence, P divides C_{1} C_{2} externally in the ratio 2 : 1$

$∴$ $Coordinates of P = (6, 0)$

$Let the equation of tangent through P be$ $y = m (x - 6)$

$As it touches x^{2} + y^{2} = 4$

$∴$ $| \frac{6 m}{\sqrt{m^{2} + 1}} |$ $= 2 \Rightarrow 36 m^{2} = 4 (m^{2} + 1) \Rightarrow m = \pm \frac{1}{2 \sqrt{2}}$

$∴$ $Equations of common tangents are$

$y = \pm \frac{1}{2 \sqrt{2}} (x - 6)$

$Also x = 2 is a common tangent to the two circles.$

Solution

Q.
A tangent $P T$ is drawn to the circle $x^{2} + y^{2} = 4$ at the point $P (\sqrt{3}, 1)$ . A straight line $L$ , perpendicular to $P T$ , is a tangent to the circle ${(x - 3)}^{2} + y^{2} = 1$ . [2012]

Q. A common tangent of the two circles is

1 $x = 4$

2 $y = 2$

3 $x + \sqrt{3} y = 4$

4 $x + 2 \sqrt{2} y = 6$

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Solution

Q. A tangent PT is drawn to the circle x2+y2=4 at the point P(3,1). A straight line L, perpendicular to PT, is a tangent to the circle (x-3)2+y2=1. [2012] Q. A common tangent of the two circles is

1 x=4

2 y=2

3 x+3y=4