Find-the-equations-of-the-circles-passing-through-4-3-and-touching-the-lines-x-y-2-and-x-y-2-

Question Number 144201 by bramlexs22 last updated on 23/Jun/21

Find the equations of the circles

passing through (- 4, 3) and touching

the lines x + y = 2 and x - y = 2

Answered by benjo_mathlover last updated on 23/Jun/21

let the equation of circle be

x^{2} + y^{2} + 2 gx + 2 fy + c = 0 \dots (i)

it passes through (- 4, 3) therefore

25 - 8 g + 6 f + c = 0 \dots (ii)

since circle touches the lines

{\begin{cases} x + y - 2 = 0 \\ x - y - 2 = 0 \end{cases} \Rightarrow then

∣ \frac{- g - f - 2}{\sqrt{2}} ∣=∣ \frac{- g + f - 2}{\sqrt{2}} ∣= \sqrt{g^{2} + f^{2} - c} \dots (iii)

Now ∣ \frac{- g - f - 2}{\sqrt{2}} ∣=∣ \frac{- g + f - 2}{\sqrt{2}} ∣

we get f = 0 or g = - 2

case (1) for f = 0

\Rightarrow∣ \frac{- g - 2}{\sqrt{2}} ∣= \sqrt{g^{2} - c}

\Rightarrow {(g + 2)}^{2} = 2 (g^{2} - c)

\Rightarrow g^{2} - 4 g - 2 c - 4 = 0 \dots (iv)

putting f = 0 in (ii) gives 25 - 8 g + c = 0 \dots (v)

eliminaring c between (iv) & (v)

g^{2} - 20 g + 46 = 0; g = 10 \pm 3 \sqrt{6}

substituting the value of g in (v)

we find c = 55 \pm 24 \sqrt{6}

so the eq of circle is

x^{2} + y^{2} \pm 2 (10 \pm 3 \sqrt{6}) x + (55 \pm 24 \sqrt{6}) = 0