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Intercept Made by Circle on Axes

Find the equa...

Question

Find the equation of a circle which touches both the axes and the line $3 x - 4 y + 8 = 0$ and lies in the third quadrant.

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Solution

The circle is as shown in figure.

$x_{1} = ∣ ∣ \begin{matrix} \frac{3 x_{1} - 4 y_{1} + 8}{5} \end{matrix} ∣ ∣ = y_{1}$

$5 x_{1} = - 3 x_{1} + 4 y^{1 - 8} 8 x_{1} = - 4 x_{1} = - 8 \dots (1) - 5 y_{1} = - 3 x_{1} + 4 y_{1} + 8 3 x_{1} + 4 y_{1} = 8 \dots (2) S o l v i n g (1) a n d (2) w e g e t, x_{1} = 2, y_{1} = 2$
As centre is in third quadrant, centre $(x_{1}, y_{1}) = (- 2, - 2)$
Equation of the circle with centre (-2,-2) and radius 2 is $(x + 2)^{2} + (y + 2)^{2} = 4$
$x^{2} + 4 x + 4 + y^{2} + 4 y + 4 = 4 x^{2} + y^{2} + 4 x + 4 y + 4 = 0$

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