Question

If the circles x² + y2 + kx + y = 0 and x2 + y2 +
4x - 2y = 0 touch each other then k =
​

Answer 1

SOLUTION

GIVEN

The circles

$\sf{ {x}^{2} + {y}^{2} + kx + y = 0 }$

and

$\sf{ {x}^{2} + {y}^{2} + 4x - 2y = 0 }$

touch each other

TO DETERMINE

The value of k

CONCEPT TO BE IMPLEMENTED

If two circles

$\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy = 0} \: \: and \: \: \sf{ {x}^{2} + {y}^{2} + 2g' x + 2f'y = 0}$

touches each other then $\sf{gf' = g' f}$

EVALUATION

Here the equation of first circle is

$\sf{ {x}^{2} + {y}^{2} + kx + y = 0 }$

Comparing with

$\sf{ {x}^{2} + {y}^{2} + 2gx + 2fy = 0} \: \: we \: get$

$\displaystyle \sf{g = \frac{k}{2} \: \: and \: \: f = \frac{1}{2} }$

Again the equation of second circle is

$\sf{ {x}^{2} + {y}^{2} + 4x - 2y = 0 }$

Comparing with

$\sf{ {x}^{2} + {y}^{2} + 2g' x + 2f'y = 0} \: \: we \: get$

$\sf{ g' = 2 \: \: and \: \: f' = - 1}$

Since the two circles touch each other

$\sf{gf' = g' f}$

$\implies \displaystyle \sf{ \frac{k}{2} \times - 1 = 2 \times \frac{1}{2} }$

$\implies \displaystyle \sf{ - \frac{k}{2} =1 }$

$\implies \displaystyle \sf{k = - 2}$

FINAL ANSWER

The required values of k = - 2

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