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
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
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For circle interior/exterior problems, always compute \(d^2\) instead of \(d\) to avoid square roots and speed up comparison with \(r^2\).
Updated On: Jun 8, 2026
- \( (-1,-3) \)
- \( (-3,1) \)
- \( (1,-3) \)
- \( (3,1) \)
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The Correct Option is D
Solution and Explanation
Concept:
• The intersection of two normal lines of a circle gives its center.
• A point lies inside the circle if its distance from the center is less than the radius.
• Radius is given as the perpendicular distance from a point to a line.
Step 1: Find the center of the circle.
Given lines: \[ 2x + y - 2 = 0 \quad \text{and} \quad 6x - 4y + 1 = 0 \] From first equation: \[ y = 2 - 2x \] Substitute into second: \[ 6x - 4(2 - 2x) + 1 = 0 \] \[ 6x - 8 + 8x + 1 = 0 \] \[ 14x - 7 = 0 \Rightarrow x = \frac{1}{2} \] \[ y = 2 - 2\left(\frac{1}{2}\right) = 1 \] So, center of the circle is: \[ C\left(\frac{1}{2}, 1\right) \]
Step 2: Find the radius.
Radius is 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 3: Check each option for interior condition.
A point is inside if: \[ (x - \tfrac{1}{2})^2 + (y - 1)^2 < 9 \] Check Option (B) \( (-3,1) \): \[ d^2 = \left(-3 - \frac{1}{2}\right)^2 + (1 - 1)^2 = \left(-\frac{7}{2}\right)^2 + 0 = \frac{49}{4} = 12.25 \] This is \(> 9\), so (B) is not inside. Now check Option (A) \( (-1,-3) \): \[ d^2 = \left(-1 - \frac{1}{2}\right)^2 + (-3 - 1)^2 = \left(-\frac{3}{2}\right)^2 + (-4)^2 = \frac{9}{4} + 16 = \frac{73}{4} = 18.25 > 9 \] Option (C) \( (1,-3) \): \[ d^2 = \left(1 - \frac{1}{2}\right)^2 + (-3 - 1)^2 = \left(\frac{1}{2}\right)^2 + 16 = \frac{1}{4} + 16 = \frac{65}{4} = 16.25 > 9 \] Option (D) \( (3,1) \): \[ d^2 = \left(3 - \frac{1}{2}\right)^2 + (1 - 1)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25 < 9 \] So the point inside the circle is: \[ \boxed{(3,1)} \]
• The intersection of two normal lines of a circle gives its center.
• A point lies inside the circle if its distance from the center is less than the radius.
• Radius is given as the perpendicular distance from a point to a line.
Step 1: Find the center of the circle.
Given lines: \[ 2x + y - 2 = 0 \quad \text{and} \quad 6x - 4y + 1 = 0 \] From first equation: \[ y = 2 - 2x \] Substitute into second: \[ 6x - 4(2 - 2x) + 1 = 0 \] \[ 6x - 8 + 8x + 1 = 0 \] \[ 14x - 7 = 0 \Rightarrow x = \frac{1}{2} \] \[ y = 2 - 2\left(\frac{1}{2}\right) = 1 \] So, center of the circle is: \[ C\left(\frac{1}{2}, 1\right) \]
Step 2: Find the radius.
Radius is 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 3: Check each option for interior condition.
A point is inside if: \[ (x - \tfrac{1}{2})^2 + (y - 1)^2 < 9 \] Check Option (B) \( (-3,1) \): \[ d^2 = \left(-3 - \frac{1}{2}\right)^2 + (1 - 1)^2 = \left(-\frac{7}{2}\right)^2 + 0 = \frac{49}{4} = 12.25 \] This is \(> 9\), so (B) is not inside. Now check Option (A) \( (-1,-3) \): \[ d^2 = \left(-1 - \frac{1}{2}\right)^2 + (-3 - 1)^2 = \left(-\frac{3}{2}\right)^2 + (-4)^2 = \frac{9}{4} + 16 = \frac{73}{4} = 18.25 > 9 \] Option (C) \( (1,-3) \): \[ d^2 = \left(1 - \frac{1}{2}\right)^2 + (-3 - 1)^2 = \left(\frac{1}{2}\right)^2 + 16 = \frac{1}{4} + 16 = \frac{65}{4} = 16.25 > 9 \] Option (D) \( (3,1) \): \[ d^2 = \left(3 - \frac{1}{2}\right)^2 + (1 - 1)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} = 6.25 < 9 \] So the point inside the circle is: \[ \boxed{(3,1)} \]
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