Skip to content
Miscellaneous Exercise 6 (II) · Q59

Q.Show that the circles x2+y2−4x−4y−28=0x^2+y^2-4x-4y-28=0 and x2+y2−4x−12=0x^2+y^2-4x-12=0 touch each other internally. Find their point of contact and the equation of their common tangent.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
79% · 59/75 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Circle 1: x2+y2−4x−4y−28=0x^2+y^2-4x-4y-28=0, so g1=−2,f1=−2,c1=−28g_1=-2,f_1=-2,c_1=-28; centre C1=(2,2)C_1=(2,2), r1=4+4+28=36=6r_1=\sqrt{4+4+28}=\sqrt{36}=6. Circle 2: x2+y2−4x−12=0x^2+y^2-4x-12=0, so g2=−2,f2=0,c2=−12g_2=-2,f_2=0,c_2=-12; centre C2=(2,0)C_2=(2,0), r2=4+0+12=16=4r_2=\sqrt{4+0+12}=\sqrt{16}=4. Distance C1C2=(2−2)2+(2−0)2=2=∣r1−r2∣=∣6−4∣C_1C_2=\sqrt{(2-2)^2+(2-0)^2}=2=|r_1-r_2|=|6-4|. So the circles touch internally.

Point of contact divides C1C2C_1C_2 externally in ratio r1:r2=6:4=3:2r_1:r_2=6:4=3:2: (6(2)−4(2)6−4, 6(0)−4(2)6−4)=(12−82,0−82)=(2,−4)\left(\dfrac{6(2)-4(2)}{6-4},\ \dfrac{6(0)-4(2)}{6-4}\right)=\left(\dfrac{12-8}{2},\dfrac{0-8}{2}\right)=(2,-4). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.