Question

Find the equation of the straight line which passes through
the centre of the circle x2 + y2 + 2x + 2y - 23 = 0 and is perpendicular to the straight line x-y+8=0.​

Answer 1

answer is equation

x+y+2=0

Answer 2

we have to find the equation of the straight line which passes through the centre of the circle x² + y² + 2x + 2y - 23 = 0 and is perpendicular to the straight line x - y + 8 = 0.

solution : equation of circle is x² + y² + 2x + 2y - 23 = 0.

so centre of circle is (-1, -1)

slope of line (x - y + 8 = 0) = 1

as unknown straight is perpendicular to the line x - y + 8 = 0

so slope of unknown line = -1/(slope of x - y + 8 = 0) = -1

now the equation straight line is given by (y - y₁) = m(x - x₁)

here (x₁ , y₁) = (-1, -1) and m = -1

⇒{y - (-1)} = (-1){x - (-1)}

⇒y + 1 = -(x + 1)

⇒y + 1 + x + 1 = 0

⇒x + y + 2 = 0

Therefore the equation of straight line is x + y + 2 = 0.

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