Question

Find the equation of the circle whose centre is c(-2,3) and which touches the line x-y+7=0

Answer 1

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Answer 2

$x^2 + y^2 + 4x -6y+11 = 0$

is the required equation of circle

Step-by-step explanation:

We are given the following in the question:

Center of circle = c(-2,3)

The circle touches the line x-y+7=0

The radius of circle will be the perpendicular distance between the centre of circle and given line.

Formula:

$(x_1, y_1), Ax + By +C = 0\\\\\text{Perpendicular distance} = \Bigg|\displaystyle\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}\Bigg|$

Radius of circle =

$(-2,3), x-y+7=0\\\\\text{Perpendicular distance} = \Bigg|\displaystyle\frac{-2-3+7}{\sqrt{(1)^2 + (-1)^2}}\Bigg| = \sqrt{2}$

Equation of circle:

$(x-h)^2 + (y-k)^2 = r^2$

where(h,k) is the center of the circle.

Putting the values, we get,

$(x+2)^2 + (y-3)^2 = (\sqrt{2})^2\\x^2 + 4x + 4 + y^2 + 9 - 6y = 2\\x^2 + y^2 + 4x -6y+11 = 0$

is the required equation of circle.

#LearnMore

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