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:

• A. $x^2+y^2-10x+2y+1=0$
• B. $x^2+y^2+6x-10y+9=0$
• C. $x^2+y^2+6x+2y-15=0$
• D. $x^2+y^2=5$

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Understand the geometry.
Two equal circles of radius $5$ touch each other at $P(1,2)$, and the common tangent there is $4x+3y=10$. The key fact is that the line joining the two centres passes through $P$ and is perpendicular to the tangent, so each centre sits on the normal to the tangent at $P$.
Step 2: Find the direction of the normal.
The tangent $4x+3y=10$ has normal direction along its coefficients, namely $(4,3)$. The length of this direction vector is $\sqrt{4^2+3^2}=5$, which is exactly the radius. How convenient.
Step 3: Make a unit-step toward each centre.
Since $(4,3)$ already has length $5$, stepping from $P$ by this whole vector moves a distance equal to the radius. So the two centres are $P\pm(4,3)$.
Step 4: Compute the two candidate centres.
\[ C_1=(1+4,\,2+3)=(5,5),\qquad C_2=(1-4,\,2-3)=(-3,-1). \]
Step 5: Write each circle and match the options.
For $C_1=(5,5)$: $(x-5)^2+(y-5)^2=25$, i.e. $x^2+y^2-10x-10y+25=0$, which is not listed.
For $C_2=(-3,-1)$: $(x+3)^2+(y+1)^2=25$. Expanding, \[ x^2+y^2+6x+2y+10=25. \]
Step 6: Simplify to standard form.
Bringing $25$ over gives $x^2+y^2+6x+2y-15=0$, which is exactly one of the options. So this is the required circle.
\[ \boxed{x^2+y^2+6x+2y-15=0} \]