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Circles Part 2

Section: Mathematics

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Question : 87 of 87

Marks: +1, -0

Let a circle C touch the lines

L_1: 4x-3y+K_1=0

L_2=4x-3y+K_2=0, K_1, K_2 \in R

. If a line passing through the centre of the circle

C

L_1

(-1,2)

L_2

(3,-6)

, then the equation of the circle

C

[25-Jun-2022-Shift-1]

$(x-1)^2+(y-2)^2=4$
$(x+1)^2+(y-2)^2=4$
$(x-1)^2+(y+2)^2=16$
$(x-1)^2+(y-2)^2=16$

Solution: 👈: Video Solution

Co-ordinate of centre

C \equiv \left( \frac{3+(-1)}{2}, \; \frac{-6-2}{2} \right) \equiv (1,-2)

L_1

is passing through

A

\Rightarrow -4-6+K_1=0

\Rightarrow K_1=10

L_2

is passing through

B

\Rightarrow 12+18+K_2=0

\Rightarrow K_2=-30

Equation of

L_1: 4x-3y+10=0

Equation of

L_2=4x-3y-30=0

Diameter of circle

=\left| \; \frac{10+30}{\sqrt{4^2+(-3)^2}} \right|=8

\Rightarrow

Radius

=4

Equation of circle

(x-1)^2+(y+2)^2=16

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