If the smallest radius of a circle passing through the intersection of x2+y2+2 x=0 and x-y=0, is r then the value of (10 r2) is equal to

Question

If the smallest radius of a circle passing through the intersection of x2+y2+2 x=0 and x-y=0, is r then the value of (10 r2) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

On solving x2+y2+2 x=0 and x-y=0, we get A(0,0) and B(-1,-1) Clearly, required circle is the circle described on A B as diameter. So radius =r=(A B/2)=(√2/2)=(1/√2). <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/3dc84a1dbfd4bff23520153f92a8714e-.png" /> Hence (10 r2)=10((1/2))=5

Q. If the smallest radius of a circle passing through the intersection of $x^{2} + y^{2} + 2 x = 0$ and $x - y = 0$ , is $r$ then the value of $(10 r^{2})$ is equal to

On solving $x^{2} + y^{2} + 2 x = 0$ and $x - y = 0$ , we get
$A (0, 0)$ and $B (- 1, - 1)$
Clearly, required circle is the circle described on $A B$ as diameter.
So radius $= r = \frac{A B}{2} = \frac{2}{2} = \frac{1}{2}$ .

Hence $(10 r^{2}) = 10 (\frac{1}{2}) = 5$

Q. If the smallest radius of a circle passing through the intersection of x2+y2+2x=0 and x−y=0, is r then the value of (10r2) is equal to

On solving x2+y2+2x=0 and x−y=0, we get A(0,0) and B(−1,−1) Clearly, required circle is the circle described on AB as diameter. So radius =r=2AB​=22​​=2​1​. Hence (10r2)=10(21​)=5

Q. If the smallest radius of a circle passing through the intersection of $x^{2} + y^{2} + 2 x = 0$ and $x - y = 0$ , is $r$ then the value of $(10 r^{2})$ is equal to

On solving $x^{2} + y^{2} + 2 x = 0$ and $x - y = 0$ , we get
$A (0, 0)$ and $B (- 1, - 1)$
Clearly, required circle is the circle described on $A B$ as diameter.
So radius $= r = \frac{A B}{2} = \frac{2}{2} = \frac{1}{2}$ .

Hence $(10 r^{2}) = 10 (\frac{1}{2}) = 5$