Question

If \( 2x+y-2=0 \) and \( 6x-4y+1=0 \) are two normals of a circle S and the length of the perpendicular drawn from (2, 3) to the line \( 3x+4y-3=0 \) is the radius of S, then the interior point of the circle S among the following options is

• A. \( (-1,-3) \)
• B. \( (-3,1) \)
• C. \( (1,-3) \)
• D. \( (3,1) \)

Hint:

For circle interior/exterior problems, always compute \(d^2\) instead of \(d\) to avoid square roots and speed up comparison with \(r^2\).

Answer 1

The Correct Option is D

Step 1: Use the property of normals.
Every normal of a circle passes through its centre. So the centre is exactly where the two given normal lines cross.
Step 2: Find the centre.
The normals are $2x+y-2=0$ and $6x-4y+1=0$. From the first, $y=2-2x$. Put this into the second: \[ 6x-4(2-2x)+1=0\implies 14x-7=0\implies x=\tfrac{1}{2} \] Then $y=2-2(\tfrac{1}{2})=1$. So the centre is $C\left(\tfrac{1}{2},1\right)$.
Step 3: Find the radius.
The radius equals the perpendicular distance from $(2,3)$ to the line $3x+4y-3=0$: \[ r=\frac{|3(2)+4(3)-3|}{\sqrt{3^2+4^2}}=\frac{|6+12-3|}{5}=\frac{15}{5}=3 \] So $r^2=9$.
Step 4: Set up the interior test.
A point lies inside the circle when its squared distance from the centre is less than $r^2=9$: \[ \left(x-\tfrac{1}{2}\right)^2+(y-1)^2<9 \]
Step 5: Test the options quickly.
For $(-3,1)$: $\left(-\tfrac{7}{2}\right)^2+0=12.25>9$ (outside). For $(-1,-3)$: $\tfrac{9}{4}+16=18.25>9$ (outside). For $(1,-3)$: $\tfrac{1}{4}+16=16.25>9$ (outside).
Step 6: Check the last point.
For $(3,1)$: $\left(\tfrac{5}{2}\right)^2+0=\tfrac{25}{4}=6.25<9$, so this point is inside. \[ \boxed{(3,1)} \]