Question:
Two circles, each of radius 5, touch at $(1,2)$. If the common tangent at the point of contact is $4x+3y=10$, then the equation of one of the circles is:
Two circles, each of radius 5, touch at $(1,2)$. If the common tangent at the point of contact is $4x+3y=10$, then the equation of one of the circles is:
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Whenever a circle touches a line at a known point, the centre lies on the normal to the line through that point. If the radius is known, simply move a distance equal to the radius along the normal direction to obtain the centre.
Updated On: Jun 9, 2026
- $x^2+y^2-10x+2y+1=0$
- $x^2+y^2+6x-10y+9=0$
- $x^2+y^2+6x+2y-15=0$
- $x^2+y^2=5$
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The Correct Option is C
Solution and Explanation
Concept:
If two circles touch externally at a point, then:
• The common tangent at the point of contact is perpendicular to the line joining the centres.
• The centre of each circle lies on the normal drawn to the tangent at the point of contact.
• The distance from the point of contact to the centre equals the radius.
Step 1: Find the normal to the tangent. The common tangent is \[ 4x+3y=10 \] A normal to this line has direction ratios \[ (4,3) \] Since the radius is \(5\), \[ \sqrt{4^2+3^2}=5 \] Hence \((4,3)\) is already a vector of length \(5\).
Step 2: Locate the centres of the circles. The circles touch at \[ P(1,2) \] Moving a distance \(5\) along the normal gives \[ C_1=(1+4,\;2+3)=(5,5) \] and in the opposite direction \[ C_2=(1-4,\;2-3)=(-3,-1) \] Thus the two possible centres are \[ (5,5)\quad \text{and}\quad (-3,-1) \]
Step 3: Form the equations of the circles. For centre \((5,5)\), \[ (x-5)^2+(y-5)^2=25 \] \[ x^2+y^2-10x-10y+25=0 \] which is not among the options. For centre \((-3,-1)\), \[ (x+3)^2+(y+1)^2=25 \] Expanding, \[ x^2+y^2+6x+2y+10=25 \] \[ x^2+y^2+6x+2y-15=0 \] Therefore, the required circle is \[ \boxed{x^2+y^2+6x+2y-15=0} \]
• The common tangent at the point of contact is perpendicular to the line joining the centres.
• The centre of each circle lies on the normal drawn to the tangent at the point of contact.
• The distance from the point of contact to the centre equals the radius.
Step 1: Find the normal to the tangent. The common tangent is \[ 4x+3y=10 \] A normal to this line has direction ratios \[ (4,3) \] Since the radius is \(5\), \[ \sqrt{4^2+3^2}=5 \] Hence \((4,3)\) is already a vector of length \(5\).
Step 2: Locate the centres of the circles. The circles touch at \[ P(1,2) \] Moving a distance \(5\) along the normal gives \[ C_1=(1+4,\;2+3)=(5,5) \] and in the opposite direction \[ C_2=(1-4,\;2-3)=(-3,-1) \] Thus the two possible centres are \[ (5,5)\quad \text{and}\quad (-3,-1) \]
Step 3: Form the equations of the circles. For centre \((5,5)\), \[ (x-5)^2+(y-5)^2=25 \] \[ x^2+y^2-10x-10y+25=0 \] which is not among the options. For centre \((-3,-1)\), \[ (x+3)^2+(y+1)^2=25 \] Expanding, \[ x^2+y^2+6x+2y+10=25 \] \[ x^2+y^2+6x+2y-15=0 \] Therefore, the required circle is \[ \boxed{x^2+y^2+6x+2y-15=0} \]
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