If a circle S passes through the origin and m | vedclass

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

Question

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0,0)$,we have $c = 0$. The equatio

Vedclass · Accepted Answer

$y^2+4x=8$

Vedclass · Answer

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0,0)$,we have $c = 0$. The equation becomes $x^2 + y^2 + 2gx + 2fy = 0$. The centre of the circle is $(-g, -f)$. The circle intersects the line $x=2$ at points where $2^2 + y^2 + 2g(2) + 2fy = 0$,which simplifies to $y^2 + 2fy + (4 + 4g) = 0$. Let the roots of this quadratic equation be $y_1$ and $y_2$. The length of the intercept is $|y_1 - y_2| = 4$. Using the property $|y_1 - y_2| = \sqrt{(y_1+y_2)^2 - 4y_1y_2}$,we get $4 = \sqrt{(-2f)^2 - 4(4+4g)}$. Squaring both sides,$16 = 4f^2 - 16 - 16g$,which simplifies to $32 = 4f^2 - 16g$,or $8 = f^2 - 4g$. Since the centre is $(h, k) = (-g, -f)$,we have $g = -h$ and $f = -k$. Substituting these into the equation $8 = f^2 - 4g$,we get $8 = (-k)^2 - 4(-h)$,which is $k^2 + 4h = 8$. Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $y^2 + 4x = 8$.

If a circle $S$ passes through the origin and makes an intercept of length $4$ units on the line $x=2$,then the equation of the curve on which the centre of $S$ lies is

Similar Questions

Let $x$ and $y$ be real numbers satisfying the equation $x^2 - 4x + y^2 + 3 = 0$. If the maximum and minimum values of $x^2 + y^2$ are $M$ and $m$ respectively,then the numerical value of $M - m$ is:

$A$ point moves in such a way that the sum of the squares of its distances from the points $A(2, 0)$ and $B(-2, 0)$ is always equal to the square of the distance between $A$ and $B$. The locus of the point is

$A$ point moves in such a way that its distance from the origin is always $4$. Then the locus of the point is:

Let a variable line passing through the centre of the circle $x^2+y^2-16x-4y=0$ meet the positive coordinate axes at points $A$ and $B$. Then the minimum value of $OA+OB$,where $O$ is the origin,is equal to

The locus of the centre of circles which pass through $(0, 1)$ and touch the line $y = x$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module