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Analytical Geometry> Find the equations of the circles whose c...

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Find the equations of the circles whose centers lie on the line 4x + 3y = 2 and which touch the lines x + y + 4 = 0 and 7x - y + 4 = 0

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Utkarsh , 10 Years ago

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3 Answers

Ravi

Last Activity: 10 Years ago

This will give 2 values of theta. Hence, 2 values of coordinates of the centre. Use the length of prependicular to find the 2 corresponding radii. This should give the 2 eqns of circles satisfying the conditions mentioned.

Utkarsh

Last Activity: 10 Years ago

Thank you so much Sir!!!

Shrutik

Last Activity: 7 Years ago

Let the centre of the circle be C (h,k).

Since the Centre lies on the line 4x+3y=2, we get 4h+3k=2

h = ( 2 – 3k ) / 4 …... Eqn (1)

Since the lines 7x-y+4=0 and x+y+4 are touching the circles ( ie they are tangents ), let’s use the formula : Distance of point C(h,k) from these lines to get the radius as :

$\displaystyle \left | 7h-k+4 \right | / \sqrt{7*7+1*1}= \left | h+k+4 \right|/\sqrt{1*1+1*1}$ …...Eqn (2)

Substitute Eqn 1 in Eqn 2 to solve for k, giving k = -2 … thus h = 2 for +ve sign of Modulus and k = 6 and h = -4 for -ve sign of modulus.

Thus, the 2 centres are C₁(2,-2) and C₂(-4,6). Radii of these 2 circles are 4/sqrt(2) and 6/sqrt(2) respectively.

Finally, using Centre -radius form, we get the 2 equations as

x² + y² -4x+4y = 0 and x²+y²+8x-12y+34=0

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