Question:
The circle passing through \((1, -2)\) and touching the axis of \(x\) at \((3, 0)\) also passes through the point :
The circle passing through \((1, -2)\) and touching the axis of \(x\) at \((3, 0)\) also passes through the point :
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A circle touching the x-axis at (3, 0) has its centre on the line x = 3 and radius equal to the absolute y-coordinate of the centre.
Updated On: Nov 13, 2026
- \((-5, 2)\)
- \((2, -5)\)
- \((5, -2)\)
- \((-2, 5)\)
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The Correct Option is C
Solution and Explanation
Touching the X-axis at \((3,0)\) means centre \(= (3, k)\) and radius \(= |k|\).
Circle: \((x-3)^2 + (y-k)^2 = k^2\).
Passing through \((1,-2)\): \(4 + (-2-k)^2 = k^2 \Rightarrow 4 + 4 + 4k + k^2 = k^2 \Rightarrow k = -2\).
So the circle is \((x-3)^2 + (y+2)^2 = 4\).
Test the options: \((5,-2)\): \(4 + 0 = 4\), satisfied. \((-5,2)\): \(64+16 \ne 4\); \((2,-5)\): \(1+9 \ne 4\); \((-2,5)\): \(25+49 \ne 4\).
Answer: (3) \((5, -2)\)
Circle: \((x-3)^2 + (y-k)^2 = k^2\).
Passing through \((1,-2)\): \(4 + (-2-k)^2 = k^2 \Rightarrow 4 + 4 + 4k + k^2 = k^2 \Rightarrow k = -2\).
So the circle is \((x-3)^2 + (y+2)^2 = 4\).
Test the options: \((5,-2)\): \(4 + 0 = 4\), satisfied. \((-5,2)\): \(64+16 \ne 4\); \((2,-5)\): \(1+9 \ne 4\); \((-2,5)\): \(25+49 \ne 4\).
Answer: (3) \((5, -2)\)
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