. The shortest distance from the point (2, -7) to circle `x^2+y^2-14x-

1.	. The shortest distance from the point (2, -7) to circle `x^2+y^2-14x-10y-151=0`
Answer» False Given circle is `x^(2)+y^(2)-14x-10y-151=0` `therefore` Centre=(7,5) and Radius=`sqrt(49+25+151)=sqrt(225)=15` So, the distances between th point (2,-7) and centre of the circle is given by `d_(1)=sqrt((2-7)^(2)+(-7-5)^(2))` `=sqrt(25+144)=sqrt(169)=13` `therefore` Shortest distance,d=`abs(13-15)=2`

Discussion

The slope of line which belongs to family (1+ l) x + (1-l)y + 2(1-l) = 0 and makes shortest intercept on `x^2 = 4y - 4`
. The shortest distance from the point (2, -7) to circle `x^2+y^2-14x-10y-151=0`
A tangent to a hyperbola `x^2/a^2 - y^2/b^2 = 1` intercepts a length of unity from each of the coordinate axes, then the point `(a, b)` lies on rectangular hyperbola
If the focus of a parabola is (0,-3) and its directrix is y=3, then its equation isA. `x^(2)=-12y`B. `x^(2)=12y`C. `y^(2)=-12x`D. `y^(2)=12x`
Find the equation of the circle which circumscribes the triangle formedby the line: ` y=x+2,3y=4x a n d 2y=3x`
Find the equation of the parabola whose focus is at (-1,-2) and thedirectrix the line `x-2y+3=0`
Find the equation of the circle having `(1,-2)`as its centre and passing through the intersectionof the lines `3x+y=14a d n2x+5y=18.`
Find the equations of the hyperbola satisfying the given conditions :Foci `(0,+-sqrt(10))`, passing through (2,3)
Find the equation of the circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line `y = x-1`
The equation of the hyperbola with vertices at (0,`pm`6) and eccentricity 5/3 is….. And its foci are …….