Question

Find the equations of the circles passing through (1, -1)and touching the lines
4x + 3y + 5 = 0 and 3x – 4y – 10 = 0

Answer 1

the equations of the circles passing through (1, -1)and touching the lines

4x + 3y + 5 = 0 and 3x – 4y – 10 = 0 are x^2+y^2-2x-4y-4 = 0 and x^2 + y^2 -(13/25) x + (68/25) y + (44/25) = 0

• Given
• the circles pass through (1, -1)
• Let the equation of circle be,
• ⇒ x^2+y^2+2gx+2fy+c=0
• since, the circles pass through (1, -1), we have,
• 1^2+(-1)^2+2g(1)+2f(-1)+c=0
• ∴ 2g-2f+c+2=0

• given lines, 4x + 3y + 5 = 0 and 3x – 4y – 10 = 0
• given, the center C(-g,-f) and (1, -1) lie on the same side of the lines, we have,
• $\dfrac{-4g-3f+5}{\sqrt{3^2+4^2} } = \dfrac{3g-4f+10}{\sqrt{3^2+4^2} } = R = \sqrt{(-g-1)^2+(-f+1)^2}$

• but, we have, f = 7g + 5
• upon solving the above equations, we get,
• $25g^2 + 38g +13 =0\\\\g = -1, \dfrac{-13}{25}\\\\f = -2, \dfrac{34}{25}$
• and
• $c = -4, \dfrac{44}{25}$

• substituting all the above obtained equations in to the standard equation of circle, we get,
• x^2+y^2-2x-4y-4 = 0
• and
• x^2+y^2-(13/25)x+(68/25)y+(44/25) = 0

Answer 2

Answer:

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