If the inverse point of the point P(3, 3) wit | vedclass

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

Question

(C) The given equation of the circle is $x^2+y^2-4x+4y+4=0$.Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=4$.The center $O$ is $(-g,

Vedclass · Accepted Answer

-$4$

Vedclass · Answer

(C) The given equation of the circle is $x^2+y^2-4x+4y+4=0$.Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=4$.The center $O$ is $(-g, -f) = (2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{4+4-4} = 2$.The inverse point $Q(a, b)$ of $P(3, 3)$ lies on the line $OP$. The slope of $OP$ is $m = \frac{3 - (-2)}{3 - 2} = \frac{5}{1} = 5$.The equation of line $OP$ is $y - (-2) = 5(x - 2)$,which simplifies to $y = 5x - 12$.Since $Q(a, b)$ lies on this line,$b = 5a - 12$ (Equation $1$).By the property of inverse points,$O, Q, P$ are collinear and $OQ \cdot OP = r^2$.$OP = \sqrt{(3-2)^2 + (3-(-2))^2} = \sqrt{1^2 + 5^2} = \sqrt{26}$.$OQ = \sqrt{(a-2)^2 + (b+2)^2} = \sqrt{(a-2)^2 + (5a-12+2)^2} = \sqrt{(a-2)^2 + (5a-10)^2} = \sqrt{(a-2)^2 + 25(a-2)^2} = \sqrt{26(a-2)^2} = |a-2|\sqrt{26}$.Given $OQ \cdot OP = r^2$,we have $|a-2|\sqrt{26} \cdot \sqrt{26} = 2^2 = 4$.$26|a-2| = 4 \Rightarrow |a-2| = \frac{4}{26} = \frac{2}{13}$.Since $Q$ lies on the same side of $O$ as $P$,$a-2 = \frac{2}{13} \Rightarrow a = 2 + \frac{2}{13} = \frac{28}{13}$.Then $b = 5(\frac{28}{13}) - 12 = \frac{140 - 156}{13} = -\frac{16}{13}$.Finally,$a+5b = \frac{28}{13} + 5(-\frac{16}{13}) = \frac{28 - 80}{13} = -\frac{52}{13} = -4$.

If the inverse point of the point $P(3, 3)$ with respect to the circle $x^2+y^2-4x+4y+4=0$ is $Q(a, b)$,then $a+5b=$

Similar Questions

Two parallel chords of a circle of radius $2$ are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre,angles of $\frac{\pi}{k}$ and $\frac{2 \pi}{k}$,where $k>0$,then the value of $[k]$ is [Note : $[k]$ denotes the largest integer less than or equal to $k$ ].

Suppose $S_1$ and $S_2$ are two unequal circles,$AB$ and $CD$ are the direct common tangents to these circles. $A$ transverse common tangent $PQ$ cuts $AB$ in $R$ and $CD$ in $S$. If $AB=10$,then $RS$ is

If the straight lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are the tangents to the same circle,then the area of that circle (in square units) is

If the area of the circum-circle of the triangle formed by the line $2x + 5y + \alpha = 0$ and the positive coordinate axes is $\frac{29\pi}{4}$ sq. units,then $|\alpha| =$

Let $A(2, 3)$,$B(4, 5)$ and let $C = (x, y)$ be a point such that $(x - 2)(x - 4) + (y - 3)(y - 5) = 0$. If the area of $\Delta ABC = \sqrt{2} \text{ sq. unit}$,then the maximum number of positions of $C$ in the $xy$ plane is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module