Question:
The shortest distance from the line \(3x + 4y = 25\) to the circle \(x^2 + y^2 - 6x + 8y = 0\) is
The shortest distance from the line \(3x + 4y = 25\) to the circle \(x^2 + y^2 - 6x + 8y = 0\) is
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Shortest distance from a line to a circle is \(|\text{distance from center to line} - \text{radius}|\).
Updated On: Jul 18, 2026
- \(\frac{9}{5}\)
- \(\frac{7}{5}\)
- \(\frac{8}{5}\)
- \(\frac{13}{5}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find circle center and radius.
Circle: \(x^2 + y^2 - 6x + 8y = 0\)
Complete squares: \((x-3)^2 + (y+4)^2 = 25\)
Center \((3,-4)\), radius \(r = 5\)
Step 2: Distance from center to line.
Line: \(3x + 4y - 25 = 0\)
Distance \(d = \frac{|3\cdot3 + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 -16 -25|}{5} = \frac{32}{5}\)
Let's check carefully:
\(\text{Distance } d = \frac{|3*3 + 4*(-4) -25|}{\sqrt{3^2+4^2}} = \frac{|9 -16 -25|}{5} = \frac{|-32|}{5} = \frac{32}{5}\)
but correct shortest distance = distance from circle edge: \(D = |d - r| = |32/5 - 5| = |32/5 -25/5| = 7/5\)
Step 3: Final conclusion.
Shortest distance from line to circle is \[ \boxed{\frac{7}{5}} \]
Circle: \(x^2 + y^2 - 6x + 8y = 0\)
Complete squares: \((x-3)^2 + (y+4)^2 = 25\)
Center \((3,-4)\), radius \(r = 5\)
Step 2: Distance from center to line.
Line: \(3x + 4y - 25 = 0\)
Distance \(d = \frac{|3\cdot3 + 4(-4) - 25|}{\sqrt{3^2 + 4^2}} = \frac{|9 -16 -25|}{5} = \frac{32}{5}\)
Let's check carefully:
\(\text{Distance } d = \frac{|3*3 + 4*(-4) -25|}{\sqrt{3^2+4^2}} = \frac{|9 -16 -25|}{5} = \frac{|-32|}{5} = \frac{32}{5}\)
but correct shortest distance = distance from circle edge: \(D = |d - r| = |32/5 - 5| = |32/5 -25/5| = 7/5\)
Step 3: Final conclusion.
Shortest distance from line to circle is \[ \boxed{\frac{7}{5}} \]
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