Question

If a circle \(S\) passing through the points \(A(1, 2)\) and \(B(2, 1)\) has its centre \(C\) located in the third quadrant at a distance of \(\frac{7}{\sqrt{2}}\) units from \(AB\), then the point \(P(1, -2)\)

• A. lies inside the circle \(S\)
• B. lies outside the circle \(S\)
• C. lies on the circle \(S\)
• D. lies on the line \(AB\)

Hint:

Use perpendicular bisector and distance formula to determine center and radius, then test point's distance from center.

Answer 1

The Correct Option is A

Line \(AB\) has slope \(m = \frac{1 - 2}{2 - 1} = -1\), so its equation is: \[ y - 2 = -1(x - 1) \Rightarrow x + y = 3 \] Distance of centre \(C\) from this line is \(\frac{7}{\sqrt{2}}\), and \(C\) lies in the third quadrant, say at coordinates \((a, b)\). Use distance from a point to a line: \(\frac{|a + b - 3|}{\sqrt{2}} = \frac{7}{\sqrt{2}} \Rightarrow |a + b - 3| = 7\) So, \(a + b = -4\) (only valid in 3rd quadrant) Now, center lies on the perpendicular bisector of \(AB\). Midpoint of \(AB = (1.5, 1.5)\), and slope of perpendicular = 1 So equation: \(y - 1.5 = 1(x - 1.5) \Rightarrow y = x\) Solving: \(a + b = -4\) and \(b = a\) ⇒ \(2a = -4 \Rightarrow a = -2, b = -2\) So center = \((-2, -2)\), radius = distance to \(A(1,2)\) = \(\sqrt{(3)^2 + (4)^2} = 5\) Now, check if \(P(1, -2)\) lies inside the circle: Distance to center = \(\sqrt{(1 + 2)^2 + (0)^2} = 3<5\) Hence, point lies inside the circle.