Question:
The least distance of the point \((10,7)\) from the circle
\[
x^2+y^2-4x-2y-20=0
\]
is
The least distance of the point \((10,7)\) from the circle
\[
x^2+y^2-4x-2y-20=0
\]
is
Show Hint
For an external point, the least distance from the circle is:
\[
\text{distance from centre}-\text{radius}.
\]
Updated On: Jun 22, 2026
- \(6\)
- \(7\)
- \(4\)
- \(5\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Write the circle in standard form.
Given, \[ x^2+y^2-4x-2y-20=0 \] Complete the squares: \[ (x^2-4x)+(y^2-2y)=20 \] \[ (x-2)^2-4+(y-1)^2-1=20 \] \[ (x-2)^2+(y-1)^2=25 \]
Step 2: Identify centre and radius.
The centre is \[ C=(2,1) \] and radius is \[ r=5 \]
Step 3: Find distance of point from centre.
Let \[ P=(10,7) \] Then, \[ CP=\sqrt{(10-2)^2+(7-1)^2} \] \[ =\sqrt{8^2+6^2} \] \[ =\sqrt{64+36} \] \[ =10 \]
Step 4: Find least distance from point to circle.
Since point lies outside the circle, \[ \text{least distance}=CP-r \] \[ =10-5 \] \[ =5 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{5} \]
Given, \[ x^2+y^2-4x-2y-20=0 \] Complete the squares: \[ (x^2-4x)+(y^2-2y)=20 \] \[ (x-2)^2-4+(y-1)^2-1=20 \] \[ (x-2)^2+(y-1)^2=25 \]
Step 2: Identify centre and radius.
The centre is \[ C=(2,1) \] and radius is \[ r=5 \]
Step 3: Find distance of point from centre.
Let \[ P=(10,7) \] Then, \[ CP=\sqrt{(10-2)^2+(7-1)^2} \] \[ =\sqrt{8^2+6^2} \] \[ =\sqrt{64+36} \] \[ =10 \]
Step 4: Find least distance from point to circle.
Since point lies outside the circle, \[ \text{least distance}=CP-r \] \[ =10-5 \] \[ =5 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{5} \]
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