Question:
The equation of the line that is a tangent to the circle
\[
x^2+y^2-6x+4y+12=0
\]
is
The equation of the line that is a tangent to the circle
\[
x^2+y^2-6x+4y+12=0
\]
is
Show Hint
To test whether a line is tangent to a circle:
\[
\text{Distance from centre to line}
=
\text{Radius}.
\]
Compute the centre and radius first, then use the point-to-line distance formula.
Updated On: Jul 29, 2026
- \[ x+3y+7=0 \]
- \[ 12x+5y+11=0 \]
- \[ 5x+12y-4=0 \]
- \[ x-3y-7=0 \]
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
A line is tangent to a circle if the perpendicular distance from the centre of the circle to the line is equal to the radius of the circle.
For a circle
\[
x^2+y^2+2gx+2fy+c=0,
\]
its centre is
\[
(-g,-f)
\]
and radius is
\[
r=\sqrt{g^2+f^2-c}.
\]
Step 1: Find the centre and radius of the circle. Given \[ x^2+y^2-6x+4y+12=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-3, \qquad f=2, \qquad c=12. \] Hence the centre is \[ (3,-2). \] The radius is \[ r = \sqrt{(-3)^2+(2)^2-12}. \] \[ = \sqrt{9+4-12}. \] \[ = 1. \]
Step 2: Check option (A). For \[ x+3y+7=0, \] distance from \((3,-2)\) is \[ d = \frac{|3-6+7|} {\sqrt{1^2+3^2}} = \frac{4}{\sqrt{10}} \neq 1. \] Hence, not a tangent.
Step 3: Check option (B). For \[ 12x+5y+11=0, \] \[ d = \frac{|36-10+11|} {\sqrt{12^2+5^2}} = \frac{37}{13} \neq 1. \] Hence, not a tangent.
Step 4: Check option (C). For \[ 5x+12y-4=0, \] \[ d = \frac{|15-24-4|} {\sqrt{5^2+12^2}} = \frac{13}{13}. \] \[ d=1. \] Since \[ d=r, \] the line is a tangent to the circle.
Step 5: Verify remaining option. For \[ x-3y-7=0, \] \[ d = \frac{|3+6-7|} {\sqrt{1+9}} = \frac{2}{\sqrt{10}} \neq 1. \] Hence, not a tangent.
Step 6: Write the final answer. \[ \boxed{5x+12y-4=0} \]
Step 1: Find the centre and radius of the circle. Given \[ x^2+y^2-6x+4y+12=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-3, \qquad f=2, \qquad c=12. \] Hence the centre is \[ (3,-2). \] The radius is \[ r = \sqrt{(-3)^2+(2)^2-12}. \] \[ = \sqrt{9+4-12}. \] \[ = 1. \]
Step 2: Check option (A). For \[ x+3y+7=0, \] distance from \((3,-2)\) is \[ d = \frac{|3-6+7|} {\sqrt{1^2+3^2}} = \frac{4}{\sqrt{10}} \neq 1. \] Hence, not a tangent.
Step 3: Check option (B). For \[ 12x+5y+11=0, \] \[ d = \frac{|36-10+11|} {\sqrt{12^2+5^2}} = \frac{37}{13} \neq 1. \] Hence, not a tangent.
Step 4: Check option (C). For \[ 5x+12y-4=0, \] \[ d = \frac{|15-24-4|} {\sqrt{5^2+12^2}} = \frac{13}{13}. \] \[ d=1. \] Since \[ d=r, \] the line is a tangent to the circle.
Step 5: Verify remaining option. For \[ x-3y-7=0, \] \[ d = \frac{|3+6-7|} {\sqrt{1+9}} = \frac{2}{\sqrt{10}} \neq 1. \] Hence, not a tangent.
Step 6: Write the final answer. \[ \boxed{5x+12y-4=0} \]
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