GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 32 of 65

Marks: +1, -0

Suppose the circles

{x}^{2}+ {y}^{2}=1

and

( {x}-1)^{2}+( {y}-1)^{2}= {r}^{2}

intersect each other orthogonally at the point

( {u}, {v})

. Then

{u}+ {v}=

Solution:

{x}^{2}+ {y}^{2}=1

.......(1)

(x-1)^{2}+(y-1)^{2}=r^{2}

......(2)
If (u, v) lies on both the circles (1) & (2),

(1) ⇒ u^{2}+v^{2}=1

{N}↖{→}_{1}=\ov{∇} φ_{1}={i}↖{∧}(2 x)+{j}↖{∧}(2 y)=2 u {i}↖{∧}+2 v {j}↖{∧}

{N}↖{→}_{2}=\ov{∇} φ_{2}=2(x-1) i+2(y-1) {j}↖{∧}=2(u-1) {i}↖{∧}+2(v-1) {j}↖{∧}

Now,

{N}↖{→}_{1} ⋅ {N}↖{→}_{2}=0

2 {u} .2( {u}-1)+2 {v} .2( {v}-1)=0

4[u^{2}+v^{2}-u-v]=0

4[1-u-v]=0

- {u}- {v}+1=0

{u}+ {v}=1

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