Question

The equation of the circle concentric with circle \(x^2+y^2-6x+7 = 0\) and which touches the line \(x+y+3 = 0\) is ....

• A. \(x^2+y^2-6x+9 = 0\)
• B. \(x^2+y^2-6x-9 = 0\)
• C. \(x^2+y^2-6x+3 = 0\)
• D. \(x^2+y^2-6x-3 = 0\)

Hint:

Same centre \((3,0)\); radius equals the distance from the centre to the tangent line.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Concentric circles share a centre. The new circle touches the line, so its radius is the perpendicular distance from the centre to the line.

Step 2: Find the centre:
\(x^2+y^2-6x+7 = 0\) has centre \((3,0)\).

Step 3: Find the radius:
\[ r = \frac{|3 + 0 + 3|}{\sqrt{1^2+1^2}} = \frac{6}{\sqrt2} = 3\sqrt2,\quad r^2 = 18 \]

Step 4: Equation:
\((x-3)^2 + y^2 = 18\), so \(x^2 + y^2 - 6x + 9 - 18 = 0\), i.e. \(x^2+y^2-6x-9 = 0\).
Option A has radius \(0\) (a point), and C and D have radius squared \(6\) and \(12\), which do not equal \(18\).

Final Answer:
The circle is \(x^2+y^2-6x-9=0\), option (B). \[ \boxed{x^2+y^2-6x-9=0} \]