A circle S passes through the point (0,1) and | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Let the cirlce be

$x^2+y^2+2 g x+2 f y+c=0$ $\quad\quad..........(1)$

given circles

$x ^2+ y ^2-2 x -15=0 $ $\quad\quad..........(2)$

Vedclass · Accepted Answer

$(B,C)$

Vedclass · Answer

Let the cirlce be

$x^2+y^2+2 g x+2 f y+c=0$ $\quad\quad..........(1)$

given circles

$x ^2+ y ^2-2 x -15=0 $ $\quad\quad..........(2)$

$x ^2+ y ^2-1=0$  $\quad\quad..........(3)$

$(1)$ and $(2)$ are orthogonal

$\Rightarrow \quad-g+0=\frac{c-15}{2} $

$\Rightarrow \quad 0+0=\frac{c-1}{2} $

$\Rightarrow \quad c=1 \& g=7$

so the cirle is $x^2+y^2+14 x+2 f y+1=0 \quad$ it passes thrgouh

$(0,1)  \Rightarrow \quad 0+1+0+2 f+1=0 $

$ \Rightarrow \quad x^2+y^2+14 x-2 y+1=0$

Centre $(-7,1)$

$	ext { radius }=7$

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then

$(A)$ radius of $S$ is $8$

$(B)$ radius of $S$ is $7$

$(C)$ centre of $S$ is $(-7,1)$

$(D)$ centre of $S$ is $(-8,1)$

Similar Questions

Circles ${(x + a)^2} + {(y + b)^2} = {a^2}$ and ${(x + \alpha )^2}$ $ + {(y + \beta )^2} = $ ${\beta ^2}$ cut orthogonally, if

If two circles ${(x - 1)^2} + {(y - 3)^2} = {r^2}$ and ${x^2} + {y^2} - 8x + 2y + 8 = 0$ intersect in two distinct points, then

The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is

The intercept on the line $y = x$ by the circle ${x^2} + {y^2} - 2x = 0$ is $AB$ . Equation of the circle with $AB$ as a diameter is

Give the number of common tangents to circle ${x^2} + {y^2} + 2x + 8y - 23 = 0$ and ${x^2} + {y^2} - 4x - 10y + 9 = 0$

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then $(A)$ radius of $S$ is $8$ $(B)$ radius of $S$ is $7$ $(C)$ centre of $S$ is $(-7,1)$ $(D)$ centre of $S$ is $(-8,1)$