The number of real circles cutting orthogonally the circle x2 + y2 + 2x - 2y + 7 = 0 is

Question

The number of real circles cutting orthogonally the circle x2 + y2 + 2x - 2y + 7 = 0 is (A) 0 (B) 1 (C) 2 (D) infinitely many

Tardigrade Admin · Accepted Answer

Given, equation of circle is x2+y2+2 x-2 y+7=0 Here, radius of the circle =√(1)+(-1)2-7 =√1+1-7=√-5 = imaginary ∴ Given circle is an imaginary circle. Hence, number of real circles cutting orthogonally the given imaginary circle is zero.

Q. The number of real circles cutting orthogonally the circle $x^{2} + y^{2} + 2 x - 2 y + 7 = 0$ is

0

1

2

infinitely many

Given, equation of circle is
$x^{2} + y^{2} + 2 x - 2 y + 7 = 0$
Here, radius of the circle
$= (1) + (- 1)^{2} - 7$
$= 1 + 1 - 7 = - 5$
$=$ imaginary
$∴$ Given circle is an imaginary circle.
Hence, number of real circles cutting orthogonally the given imaginary circle is zero.

Q. The number of real circles cutting orthogonally the circle x2+y2+2x−2y+7=0 is

0

1

2

infinitely many

Given, equation of circle is x2+y2+2x−2y+7=0 Here, radius of the circle =(1)+(−1)2−7​ =1+1−7​=−5​ = imaginary ∴ Given circle is an imaginary circle. Hence, number of real circles cutting orthogonally the given imaginary circle is zero.

Q. The number of real circles cutting orthogonally the circle $x^{2} + y^{2} + 2 x - 2 y + 7 = 0$ is

Given, equation of circle is
$x^{2} + y^{2} + 2 x - 2 y + 7 = 0$
Here, radius of the circle
$= (1) + (- 1)^{2} - 7$
$= 1 + 1 - 7 = - 5$
$=$ imaginary
$∴$ Given circle is an imaginary circle.
Hence, number of real circles cutting orthogonally the given imaginary circle is zero.