Answer:
DC = 39
Step-by-step explanation:
From inspection of the diagram:
- EA is tangent to both circles
- DE is the radius of circle D
- CA is the radius of circle C
The tangent of a circle is always perpendicular to the radius, therefore:
DE ⊥ EA and CA ⊥ EA
As ∠DEB and ∠BAC are both 90°, then DE is parallel to CA.
Therefore, ∠DBE and ∠ABC are vertically opposite angles, and are therefore equal.
As triangles ΔBED and ΔBAC have two pairs of corresponding congruent angles, the triangles are similar.
Therefore:
[tex]\implies \sf \dfrac{DE}{CA}=\dfrac{EB}{BA}[/tex]
[tex]\implies \sf \dfrac{10}{5}=\dfrac{24}{BA}[/tex]
[tex]\implies \sf BA=12[/tex]
Using Pythagoras' Theorem for ΔBED to find DB:
[tex]\implies \sf DE^2+EB^2=DB^2[/tex]
[tex]\implies \sf 10^2+24^2=DB^2[/tex]
[tex]\implies \sf DB^2=676[/tex]
[tex]\implies \sf DB=\sqrt{676}[/tex]
[tex]\implies \sf DB=26[/tex]
Using Pythagoras' Theorem for ΔBAC to find BC:
[tex]\implies \sf CA^2+BA^2=BC^2[/tex]
[tex]\implies \sf 5^2+12^2=BC^2[/tex]
[tex]\implies \sf BC^2=169[/tex]
[tex]\implies \sf BC=\sqrt{169}[/tex]
[tex]\implies \sf BC=13[/tex]
Therefore, the distance between the center of the circles DC is:
[tex]\begin{aligned} \implies \sf DC & = \sf DB + BC\\& = \sf 26 + 13\\& = \sf 39\end{aligned}[/tex]
