Question

Find the radical centre of three circles described on the three sides 4x–7y+10 = 0, x+y–5= 0, 7x+4y–15 = 0 of a triangle as diameters.

Answer 1

We know that radical centre of 3 circles described on the three sides of a triangle as diameters is orthocenter of the triangle.

L₁ ≡ 4x–7y+10 = 0________ (1)

L₂ ≡ x+y–5 = 0___________(2)

L₃ ≡ 7x+4y–15 = 0__________(3)

Slope of line L₁ is 4/7 and slope of line L₃ = –7/4 since the product of these slopes is –1

∴L₁ and L₃ are perpendicular.

We know that orthocentre of a right angled triangle is the vertex where it has right angle. Solving (1) & (3) to get orthocentre

(1)×4+(2)×7

16x – 28y + 40 = 0

\(\frac{49x+28y-105=0}{65x-65}\)

x = 1

4–7y+10 = 0

7y+14

y = 2

∴Radical centre is (1, 2)