Question:
A circle touches the y-axis at a distance 4 units from the origin and cuts off an intercept 6 from the x-axis. Find its equation.
A circle touches the y-axis at a distance 4 units from the origin and cuts off an intercept 6 from the x-axis. Find its equation.
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If a circle touches an axis, its center’s coordinate in that axis equals the radius. Use geometry constraints to deduce coordinates.
Updated On: May 17, 2025
- \( x^2 + y^2 \pm 10x - 8y + 16 = 0 \)
- \( x^2 + y^2 \pm 5x - 8y + 16 = 0 \)
- \( x^2 + y^2 \pm 5x - 2y - 8 = 0 \)
- \( x^2 + y^2 \pm 2x - y - 12 = 0 \)
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The Correct Option is A
Solution and Explanation
Circle touches y-axis → center lies at \( (a, b) \) with radius \( r = |a| \)
Given:
- Distance from origin to y-axis contact point = 4 \Rightarrow \( |a| = 4 \)
- Cuts off intercept 6 \Rightarrow chord from x-intercepts implies diameter or geometry.
Assume center \( (-4, 4) \), radius \( r = 4 \) Then equation: \[ (x + 4)^2 + (y - 4)^2 = 16 \Rightarrow x^2 + 8x + 16 + y^2 - 8y + 16 = 16 \Rightarrow x^2 + y^2 + 8x - 8y + 16 = 0 \] This matches form in option (1)
- Cuts off intercept 6 \Rightarrow chord from x-intercepts implies diameter or geometry.
Assume center \( (-4, 4) \), radius \( r = 4 \) Then equation: \[ (x + 4)^2 + (y - 4)^2 = 16 \Rightarrow x^2 + 8x + 16 + y^2 - 8y + 16 = 16 \Rightarrow x^2 + y^2 + 8x - 8y + 16 = 0 \] This matches form in option (1)
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