The equation of the circle whose centre lies | vedclass

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Question

(A) The centre of the circle is the point of intersection of the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$.Solving these equations: $y = 3x - 4$. Su

Vedclass · Accepted Answer

$x^2 + y^2 - 2x + 2y - 47 = 0$

Vedclass · Answer

(A) The centre of the circle is the point of intersection of the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$.Solving these equations: $y = 3x - 4$. Substituting into the second equation: $x + 3(3x - 4) + 2 = 0$ $\Rightarrow x + 9x - 12 + 2 = 0$ $\Rightarrow 10x = 10$ $\Rightarrow x = 1$.Then $y = 3(1) - 4 = -1$. So,the centre is $(1, -1)$.The area of the circle is $\pi r^2 = 154$. Taking $\pi = \frac{22}{7}$,we have $\frac{22}{7} r^2 = 154$ $\Rightarrow r^2 = 154 	imes \frac{7}{22} = 7 	imes 7 = 49$ $\Rightarrow r = 7$.The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 7^2$.Expanding this: $x^2 - 2x + 1 + y^2 + 2y + 1 = 49 \Rightarrow x^2 + y^2 - 2x + 2y - 47 = 0$.

The equation of the circle whose centre lies on the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$ and has an area of $154$ square units is:

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