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  • Please,please help me - If a circle C passing through the point (4,0) touches the circle externally at the point (1,-1), th - Co-ordinate geometry - JEE Main

If a circle C passing through the point (4,0) touches the circle x^{2}+y^{2}+4x-6y=12 externally at the point (1,-1), then the radius of C is:

  • Option 1)

    2\sqrt5

  • Option 2)

    4

  • Option 3)

    5

  • Option 4)

    \sqrt57

Answers (1)

best_answer

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 

 

 

Common tangents of two circles -

When two circles touch  each other externally, there are three common tangents, two of them are direct.

\\|C_{1}C_{2}|=|r_{1}+r_{2}|\\Let,\\S_1=\left(x-x_1\right)^2+\left(y-y_1\right)^2-r_1^2=0\\S_2=\left(x-x_2\right)^2+\left(y-y_2\right)^2-r_2^2=0\\equatio\:of\:comon\:tangent\\S_1-S_2=0

 

- wherein

 

From the concept we have learnt

Equation of tangent at point (1,-1) to the circle x2+y2+4x-6y-12 = 0

x-y +2(x+1) -3(y-1) -12 = 0

\Rightarrow 3x-4y-7 = 0

Equation of circle

\because S + \lambda L = 0

( x2+y2+4x-6y-12 ) + \lambda(12-7) = 0

\lambda = -4

\therefore ( x2+y2+4x-6y-12 )-4(3x-4y-7) = 0

\Rightarrow x2+y2-8x+10y+16 = 0

Radius of circle = \sqrt{g^{2}+f^{2}-c}

= \sqrt{16+25-16} = 5


Option 1)

2\sqrt5

Option 2)

4

Option 3)

5

Option 4)

\sqrt57

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