Skip to content
Miscellaneous Exercise 6 (II) · Q57

Q.Show that the circles x2+y2−4x+10y+20=0x^2+y^2-4x+10y+20=0 and x2+y2+8x−6y−24=0x^2+y^2+8x-6y-24=0 touch each other externally. Find their point of contact and the equation of their common tangent.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
76% · 57/75 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+10y+20=0x^2+y^2-4x+10y+20=0, so g1=−2,f1=5,c1=20g_1=-2,f_1=5,c_1=20; centre C1=(2,−5)C_1=(2,-5), r1=4+25−20=9=3r_1=\sqrt{4+25-20}=\sqrt9=3. Circle 2: x2+y2+8x−6y−24=0x^2+y^2+8x-6y-24=0, so g2=4,f2=−3,c2=−24g_2=4,f_2=-3,c_2=-24; centre C2=(−4,3)C_2=(-4,3), r2=16+9+24=49=7r_2=\sqrt{16+9+24}=\sqrt{49}=7. Distance C1C2=(2+4)2+(−5−3)2=36+64=100=10=r1+r2=3+7C_1C_2=\sqrt{(2+4)^2+(-5-3)^2}=\sqrt{36+64}=\sqrt{100}=10=r_1+r_2=3+7. So the circles touch externally.

Point of contact divides C1C2C_1C_2 internally in the ratio r1:r2=3:7r_1:r_2=3:7 from C1C_1: (3(−4)+7(2)10, 3(3)+7(−5)10)=(−12+1410, 9−3510)=(15,−2610)=(15,−135)\left(\dfrac{3(-4)+7(2)}{10},\ \dfrac{3(3)+7(-5)}{10}\right)=\left(\dfrac{-12+14}{10},\ \dfrac{9-35}{10}\right)=\left(\dfrac15,-\dfrac{26}{10}\right)=\left(\dfrac15,-\dfrac{13}5\right). …

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.